\documentclass[11pt]{scrartcl}

% vim: tw=50

\usepackage{notesheader}

\begin{document}

\title{Groups}
\author{}
\date{\today}
\maketitle

\tableofcontents

\newpage
\setcounter{section}{-1}
\section{Introduction}

Book recommendations:
\begin{itemize}
    \item Algebra \& Geometry, Alan Beardon
\end{itemize}

\begin{notation*}
    $\forall$ denotes ``for all''; $\exists$ denotes
    ``there exists''; $\implies$ denotes
    ``implies''; $\therefore$ denotes
    ``therefore''; \contradiction denotes
    ``contradiction''; and $\ZZ$, $\NN$, $\QQ$,
    $\RR$ and $\CC$ denote the integers, naturals,
    rationals, reals and complex numbers
    respectively.
\end{notation*}
\newpage

\section{Basic Definitions and Examples}

\begin{definition}[Binary Operation]
    A binary operation $*$ on a set $X$ is a way
    of combining 2 elements of $X$ to
    unambiguously give another element of $X$,
    i.e. $* : X \times X \to X$.
\end{definition}

\begin{definition}[Group]
    If $G$ is a set and $*$ is a binary operation
    on $G$, then $(G, *)$ is a \emph{group} if
    the following 4 axioms hold:
    \begin{enumerate}[(i)]
        \item $x, y \in G \implies x * y \in G$
            (closure)
        \item $\exists$ an element $e \in G$
            satisfying
            \[ x * e = x = e * x \qquad \forall x
            \in G \]
            (existence of an identity)
        \item for every $x \in G$ there is a $y
            \in G$ such that
            \[ x * y = e = y * x \]
            (existence of inverses)
        \item for every $x, y, z \in G$ we have:
            \[ x * (y * z) = (x * y) * z \]
            (associative law)
    \end{enumerate}
\end{definition}

\begin{remark*}
    We can prove that $G$ has only one identity.
\end{remark*}

\begin{remark*}
    As a result, we can also prove that every
    element has only one inverse.
\end{remark*}

\noindent Both of these claims are proved in Lemma 1.

\subsection{Examples of Groups}

\begin{enumerate}[(1)]
    \item $(\ZZ, +)$, $e = 0$, $x^{-1} = -x$.
    \item $(\QQ, +)$, $(\RR, +)$
    \item $(\ZZ, -)$ is \emph{not} a group because
        associativity fails.
    \item $(\ZZ, \times)$ is \emph{not} a group
        because no inverses.
    \item $(\QQ, \times)$ is \emph{not} a group
        because $0^{-1}$ does not exist.
    \item $(Q \setminus \{0\}, \times)$
    \item $(\{ \pm 1\}, \times)$ \\
        We can write a multiplication table:
        \begin{center}
            \begin{tabular}{c|cc}
                $x$ & 1 & -1 \\
                \hline
                1 & 1 & -1 \\
                -1 & -1 & 1
            \end{tabular}
        \end{center}
        note that closure holds, $e = 1$ and
        $(-1)^{-1} = -1$.
    \item $(\{0, 1, 2\}, +_3)$
        \begin{center}
            \begin{tabular}{c|ccc}
                $+_3$ & 0 & 1 & 2 \\
                \hline
                0 & 0 & 1 & 2 \\
                1 & 1 & 2 & 0 \\
                2 & 2 & 0 & 1
            \end{tabular}
        \end{center}
        and we have $e = 0$ and $1^{-1} = 2$.
    \item $(\{e, a, b, c\}, *)$
        \begin{center}
            \begin{tabular}{c|cccc}
                $*$ & $e$ & $a$ & $b$ & $c$ \\
                \hline
                $e$ & $e$ & $a$ & $b$ & $c$ \\
                $a$ & $a$ & $e$ & $c$ & $b$ \\
                $b$ & $b$ & $c$ & $e$ & $a$ \\
                $c$ & $c$ & $b$ & $a$ & $e$
            \end{tabular}
        \end{center}
    \item ``groups are abstractions of
        symmetries'': rotations and reflections
        of an equilateral triangle are another
        example of a group.
        \begin{center}
        \begin{tsqx}
            ! size(3cm);
            ~regular A B C
            A--B--C--A
            label $e$ @ (0,-1)
        \end{tsqx}
        \begin{tsqx}
            ! size(3cm);
            ~regular C A B
            A--B--C--A
            D := foot A B C
            E := 1.3*D-0.3*A
            arc3 E E+0.2*dir(-30) E+0.2*dir(90) EndArcArrow
            label $\sigma$ @ (0,-1)
        \end{tsqx}
        \begin{tsqx}
            ! size(3cm);
            ~regular B C A
            A--B--C--A
            D := foot C A B
            E := 1.3*D-0.3*C
            arc3 E E+0.2*dir(-30) E+0.2*dir(90) BeginArcArrow
            label $\sigma^2$ @ (0,-1)
        \end{tsqx}
        \end{center}
        \begin{center}
        \begin{tsqx}
            ! size(3cm);
            ~regular A C B
            A--B--C--A
            D := foot A B C
            E := 1.15*D-0.15*A
            A--E dashed
            label $\tau$ @ (0,-1)
        \end{tsqx}
        \begin{tsqx}
            ! size(3cm);
            ~regular B A C
            A--B--C--A
            D := foot C A B
            E := 1.15*D-0.15*C
            C--E dashed
            label $\tau \circ \sigma$ @ (0,-1)
        \end{tsqx}
        \begin{tsqx}
            ! size(3cm);
            ~regular C B A
            A--B--C--A
            D := foot B C A
            E := 1.15*D-0.15*B
            B--E dashed
            label $\tau \circ \sigma^2$ @ (0,-1)
        \end{tsqx}
        \end{center}
        This forms a group where the binary
        operator is ``do one then the next''
    \item $M_2(\RR) = \{2 \times 2 \text{ matrices
        with entries in $\RR$}\}$
        \[ = \left[\left(
        \begin{matrix}
            a & b \\ c & d
        \end{matrix}
        \right) : a, b, c, d \in \RR \right] \]
        under addition is a group:
        \[ \left(
        \begin{matrix}
            a & b \\ c & d
        \end{matrix}
        \right)
        + \left(
        \begin{matrix}
            \alpha & \beta \\ \gamma & \delta
        \end{matrix}
        \right) 
        = \left(
        \begin{matrix}
            a + \alpha & b + \beta \\
            c + \gamma & d + \delta
        \end{matrix}
        \right) \]
    \item $GL_2(\RR) = \{\text{invertible $2
            \times 2$ matrices with entries in
        $\RR$}\}$ under multiplication is a group.
\end{enumerate}

\begin{lemma}
    Let $(G, *)$ be a group. Then
    \begin{enumerate}
        \item The identity element is unique.
        \item Inverses are unique.
    \end{enumerate}
\end{lemma}

\begin{proof}
    (i): Suppose $e$ and $e'$ are both identities,
    so
    \[ a * e = a = e * a \qquad \text{and}
    \qquad a * e' = a = e' * a \qquad
    \forall a \in G .\]
    In particular
    \[ e = e * e' = e' \]
    so $e = e'$, so the identity must be unique.
\end{proof}

\begin{proof}
    (ii): Suppose both $y$ and $z$ are inverses
    for $x$, so
    \[ x * y = e = y * x, \qquad \text{and}
    \qquad x * z = e = z * x \qquad
    x \in G .\]
    Then
    \begin{align*}
        y &= y * e \\
          &= y * (x * z) \\
          &= (y * x) * z \\
          &= e * z \\
          &= z
    \end{align*}
    so $y = z$.
\end{proof}

\begin{remark*}[Unnesessary brackets]
    Since the definition of a group involves
    associativity, we can omit brackets, i.e.
    $x * y * z$ is unambiguous.
\end{remark*}

\begin{remark*}[Omitting $*$]
    We often omit ``$*$'' and write $xy := x * y$
    and also write $G = (G, *)$. (This is only
    done when the binary operator can be easily
    inferred).
\end{remark*}

\begin{remark*}[Inverse of product]
    $(xy)^{-1} = y^{-1}x^{-1}$. This follows
    immediately by the uniqueness, as it is easy
    to verify that this is a possible inverse:
    \[ (xy)y^{-1}x^{-1} = x(yy^{-1})x^{-1}
    = xx^{-1} = e .\]
\end{remark*}

\begin{remark*}[Inverse of inverse]
    $(x^{-1})^{-1} = x$.
\end{remark*}

\begin{remark*}[Coset stuff]
    If $xy = xz$ then $y = z$; this easily follows
    from the existence of inverses.
\end{remark*}

\begin{definition}[Abelian Groups]
    A group $G$ is \emph{abelian} (or commutative)
    if $xy = yx$ for all $x, y \in G$.
\end{definition}

\begin{remark*}
    Note all our examples above are abelian except
    (10) and (12). (Symmetries of the triangle,
    and the general linear group).
\end{remark*}

\begin{definition}[Order of a group]
    Let $G$ be a group. If the number of elements
    in the set $G$ is finite, then $G$ is called a
    \emph{finite group}. Otherwise $G$ is called
    an \emph{infinite group}. If $G$ is a finite
    group, denote the number of elements in the
    set $G$ by $|G|$ and we call this the
    \emph{order} of the group.
\end{definition}

\begin{definition}[Subgroups]
    Let $(G, *)$ be a group and $H$ a subset of
    $G$ ($H \subseteq G$ i.e. $h \in H \implies h
    \in G$). Then $(H, *)$ is a \emph{subgroup} of
    $(G, *)$ if $(H, *)$ is a group (with the same
    operation) i.e. if
    \begin{enumerate}[(a)]
        \item $h, k \in H \implies h * k \in H$.
        \item $e_G \in H$
        \item $h \in H \implies h^{-1} \in H$.
    \end{enumerate}
    (Note associativity is inherited). \\
    i.e. ``restricting operation to $H$ still
    gives a group''. We write $H \le G$.
\end{definition}

\subsubsection*{Examples}

\begin{itemize}
    \item $(\ZZ, +) \le (\QQ, +) \le (\RR, +)$
    \item $(\{\pm 1\}, \times) \le (\QQ \setminus
        \{0\}, \times)$.
    \item In example (10) (symmetries of a
        triangle), the rotational symmetries form
        a subgroup (elements $\{e, \sigma,
        \sigma^2\}$).
    \item In example (12) (general linear group),
        we have that
        \begin{align*}
            \mathrm{SL}_2(\RR) &= \{A \in
            \mathrm{GL}_2(\RR) : \det A = 1\} \\
            &\le \mathrm{GL}_2(\RR)
        \end{align*}
        ($\mathrm{SL}_2$ and $\mathrm{GL}_2$
        denote the special linear and general
        linear groups respectively).
    \item $G$ a group then $\{e\} \le G$ is the
        trivial subgroup. $G \le G$ is the
        improper subgroup.
    \item The subgroups of $(\ZZ, +)$ are exactly
        \[ n\ZZ = \{nk : k \in \ZZ\}, \qquad n \in
        \ZZ_{\ge 0} .\]
        \begin{proof}
            First note $n\ZZ$ is a sub group of
            $\ZZ$.
            \begin{itemize}
                \item $0 \in n\ZZ$
                \item If $a, b \in n\ZZ$, then let
                    $a = na'$, $b = nb'$. Then we
                    have
                    \[ a + b = na' + nb' = n(a' +
                    b') \in n\ZZ .\]
                \item $-a = n(-a') \in n\ZZ$
                \item Associativity is inherited.
            \end{itemize}
            Conversely assume that $H \le \ZZ$. If
            $H = \{0\} = 0\ZZ$ which is of the
            form we claimed. Otherwise choose $0 <
            n \in H$ with $n$ minimal. (Such an
            $n$ must exist because $H$ must
            contain either a negative or positive
            integer, but since inverses exist this
            implies that there must be a positive
            element). Then $n\ZZ \le H$ by closure
            and inverses. Now we show that $H =
            n\ZZ$. Suppose $\exists h \in H
            \setminus n\ZZ$, then we can write $h
            = nk + h'$ with $h' \in \{1, 2, \dots,
            n - 1\}$. But $h' = h - nk \in H$,
            contradicting minimality of $n$. Thus
            $H = n\ZZ$.
        \end{proof}
\end{itemize}

\subsubsection*{Definitions for Functions}

\begin{definition}[Functions]
    $F$ is a \emph{function} between sets $A$ and
    $B$ if it assigns each element of $A$ a unique
    element of $B$
     \[ f : A \to B \qquad\qquad a \mapsto f(a) \]
    For example: $f : \ZZ \to \ZZ$, $x \mapsto x +
    1$ and $g \ZZ \to \ZZ$, $x \mapsto 2x$.
\end{definition}

\begin{definition}[Composition of functions]
    Suppose $g : A \to B$ and $f : B \to C$.
    Define $f \circ g : A \to C$ by
    \[ a \mapsto (f \circ g)(a) = f(g(a)) .\]
    For example $(f \circ g)(x) = 2x + 1$ and $(g
    \circ f)(x) = 2x + 2$ using the example
    functions above.
\end{definition}

\noindent Suppose $f_1 : A \to B$, $f_2 : A \to B$. Then
$f_1 = f_2$ if and only if $f_1(a) = f_2(a)
\forall a \in A$. 

\begin{definition}[Bijection]
    $f : A \to B$ is a \emph{bijection} if it
    defines a pairing between elements of $A$ and
    elements of $B$. That is, given $b \in B$
    there exists a unique $a \in A$ such that
    $f(a) = b$. For example $f: \ZZ \to \ZZ$, $x
    \mapsto x + 1$. Given a bijective function
    $f$, we can define
    \[ f^{-1} : B \to A \qquad\qquad b \mapsto a
    \qquad \text{where $f(a) = b$} .\]
    Then $f \circ f^{-1} = \id_B$ and $f^{-1}
    \circ f = \id_A$. ($\id_B(b) = b$,
    $\id_A(a) = a$)
\end{definition}

\begin{lemma}[Composition of bijections]
    If $g : A \to B$ and $f : B \to C$ are
    bijections then so is $f \circ g : A \to C$.
\end{lemma}

\begin{proof}
    In Numbers \& Sets.
\end{proof}

\begin{definition}[Homomorphism]
    Let $(G, *_G)$ and $(H, *_H)$ be groups. Then
    the function
    \[ \theta : G \to H \]
    is a \emph{homomorphism} if
    \[ \theta(x *_G y) = \theta(x) *_H \theta(y) 
    \qquad \forall x, y \in G \]
    ``a map which respects the group operation''.
\end{definition}

\begin{example*}
    Let $G = (\{0, 1, 2, 3\}, +_4)$ and
    $H = (\{1, e^{\pi i/2}, e^{\pi i}, e^{3\pi
    i/2}\}, \times)$. Then the function
    \[ \theta : G \to H \qquad\qquad n \mapsto
    e^{n\pi i/2} \]
    is a homomorphism. This is because
    \begin{align*}
        \theta(n +_4 m) &= e^{(n +_4 m)\pi i/2} \\
        &= e^{(n + m)\pi i/2} \qquad\text{since $n
        + m = n +_4 m + 4n$} \\
        &= e^{n\pi i/2} \times e^{m\pi i/2} \\
        &= \theta(n) \times \theta(m)
    \end{align*}
\end{example*}

\begin{lemma}
    Let $G$ and $H$ be groups and
    $\theta : G \to H$ be a homomorphism. Then
    \[ \theta(G) = \{\theta(g) : g \in G\} ,\]
    the \emph{image} of $\theta$ is a subgroup of
    $H$, written $\theta(G) \le H$.
\end{lemma}

\begin{proof}
    We need to prove closure, \dots
    \begin{itemize}
        \item To prove closure, let $x, y$ be
            elements of $\theta(G)$. Then
            $x = \theta(g)$ and $y = \theta(h)$
            for some $h, g \in G$. Then:
            \begin{align*}
                x *_H y &= \theta(g) *_H \theta(h) \\
                &= \theta(g *_g h) \\
                &\in \theta(G) \qedhere
            \end{align*}
        \item To show that we have an identity,
            note that
            \begin{align*}
                \theta(e_G) &= \theta(e_G *_G e_G)
                \\
                &= \theta(e_G) *_H \theta(e_G)
            \end{align*}
            and if we premultiply by
            $\theta(e_G)^{-1} \in H$ then we get
            \[ e_H  = \theta(e_G) \in \theta(G) \]
        \item To get inverses, let $x = \theta(g)
            \in \theta(G)$. Then
            \begin{align*}
                e_H &= \theta(e_G) = \theta(g *_G
                g^{-1}) \\
                &= \theta(g) *_H \theta(g^{-1}) \\
                &= x *_H \theta(g^{-1}) \\
                &= \theta(g^{-1} *_G g) \\
                &= \theta(g^{-1}) *_H x
            \end{align*}
            And since inverses are unique we get
            \[ \theta(g)^{-1} = \theta(g^{-1}) \in
            \theta(G) \]
        \item And finally associativity is just
            inherited.
    \end{itemize}
\end{proof}

\begin{definition}[Isomorphism]
    A bijective homomorphism is called an
    \emph{isomorphism} if $G$ and $H$ are groups
    and $\theta : G \to H$ is a homomorphism. We
    say $G$ and $H$ are isomorphic and write $G
    \cong H$.
\end{definition}

\begin{example*}
    Let $G = (\{0, 1, 2, 3\}, +_4)$ and
    $H = (\{1, e^{i\pi/2}, e^{i\pi}, e^{3i\pi/2},
    \times)$. Then $G \cong H$, which can be shown
    by considering
    \begin{align*}
        \theta : G &\to H \\
        n &\mapsto e^{i\pi n/2}
    \end{align*}
    ($\theta$ is an isomorphism.)
\end{example*}
\noindent
Isomorphism means roughly ``They are essentially
the same''
\noindent
\begin{lemma}\phantom{.}
    \begin{enumerate}[(i)]
        \item The composition of two homomorphisms
            is a homomorphism. Similarly for
            isomorphisms, thus if $G_1 \cong G_2$
            and $G_2 \cong G_3$, then $G_1 \cong
            G_3$.
        \item If $\theta : G_1 \to G_2$ then so is
            its inverse $\theta^{-1} : G_2 \to
            G_1$. So $G_1 \cong G_2 \implies G_2
            \cong G_1$.
    \end{enumerate}
\end{lemma}

\begin{proof}
    \phantom{.}
    \begin{enumerate}[(i)]
        \item Suppose
            \[ \theta_1 : (G_1, *_1) \to (G_2,
            *_2) \]
            \[ \theta_2 : (G_2, *_2) \to (G_3,
            *_3) \]
            are homomorphisms. Then $\theta_2
            \circ \theta_1$ is a function from
            $G_1$ to $G_3$, we need to check its a
            homomorphism. Let $x, y \in G_1$. Then
            \begin{align*}
                \theta_2 \circ \theta_1 (x *_1 y)
                &= \theta_2(\theta_1(x) *_2
                \theta_1(y)) \\
                &= \theta_2(\theta_1(x)) *_3
                \theta_2(\theta_1(y)) \\
                &= (\theta_2 \circ \theta_1)(x)
                *_3 (\theta_2 \circ \theta_1)(y)
            \end{align*}
        \item $\theta$ is a bijection so
            $\theta^{-1}$ exists. We need to show
            it is a homomorphsim. \\
            Let $y, z \in G_2$. Then $\exists x, k
            \in G_1$ such that
            \[ \theta^{-1}(y) = x, \qquad
            \theta^{-1}(z) = k .\]
            Note
            \begin{align*}
                \theta(x *_1 k) &= \theta(x) *_2
                \theta(k) \\
                &= y *_2 z
                \implies \theta^{-1}(y *_2 z) &= x
                *_1 k \\
                &= \theta^{-1}(y) *_1
                \theta^{-1}(z)
            \end{align*}
    \end{enumerate}
\end{proof}

\begin{notation*}
    If $x \in (G, *)$, $n \in \ZZ$ then
    \[ x^n = \begin{cases}
        \overbrace{x * x * \cdots * x}^{n} & n > 0
        \\
        e & n = 0 \\
        \underbrace{x^{-1} * x^{-1} * \cdots *
        x^{-1}}_{(-n)} & n < 0
    \end{cases} \]
\end{notation*}

\begin{definition}[Cyclic Groups]
    A group $H$ is \emph{cyclic} if $\exists h
    \in H$ such that each element of $H$ is a
    power of $h$, i.e. for each $x \in H \exists n
    \in \ZZ$ such that $x = h^n$. Then $h$ is
    called a \emph{generator} of $H$ and we write
    $H = \langle h \rangle$.
\end{definition}

\begin{example*}
    \begin{itemize}
        \item $(\ZZ, +) = \langle 1 \rangle = \langle -1
        \rangle$ is the infinite cyclic group. We
        showed all subgroups of $(\ZZ, +)$ are cyclic.
        \item $(\{\pm 1\}, \times) = \langle -1
            \rangle$
        \item $(\{0, 1, 2, 3\}, +_4) = \langle 1
            \rangle = \langle 3 \rangle$
    \end{itemize}
\end{example*}

Note that a cyclic group is always abelian.

\begin{definition}[Orders]
    Let $G$ be a group and $g \in G$. The order of
    $g$ written $o(g)$, is the least positive
    integer $n$ such that $g^n = e$, if it exists.
    Otherwise $g$ has infinite order.
\end{definition}

\begin{lemma}
    Suppose $G$ is a group, $g \in G$ and $o(g) =
    m$. Let $n \in \NN_{> 0}$. Then
    \[ g^n = e \iff m \mid n .\]
\end{lemma}

\begin{proof}
    $(\Leftarrow)$ Suppose $m \mid n$, then $n =
    qm$ for some $q \in \NN$. This implies that
    \[ g^{n} = g^{qm} = (g^m)^q = e^q = e .\]
    $(\implies)$ Suppose $g^n = e$. Then we can
    write $n = qm + r$ with $0 \le r < m$, with $q
    \in \NN$. Then
    \begin{align*}
        e = g^n &= g^{qm + r} \\
        &= (g^m)^q g^r \\
        &= e^q g^r \\
        &= eg^r \\
        &= g^r
    \end{align*}
    This implies $r = 0$ by minimality of $m$,
    hence $n = qm$ as required.
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item Suppose $g \in G$> Then $\{g^n : n \in
        \ZZ\}$ is a subgroup of $G$, in fact it is
        the smallest subgroup of $G$ containing
        $g$. We call it the subgroup of $G$
        generated by $g$ and write
        \[ \langle g \rangle = \{g^n : n \in \ZZ\}
        .\]
        Also $|\langle g \rangle| = o(g)$ if
        finite, since if $o(g) = m$ then
        \[ \langle g \rangle = \{e, g, g^2, \dots,
        \underbrace{g^{m - 1}}_{=g^{-1}}\} \]
        Otherwise both are infinite.
    \item We can define the abstract cyclic group
        of order $n$
        \[ C_n = \langle x \rangle \qquad o(x) = n \]
        Then
        \[ (\{0, 1, \dots, n - 1\}, +_n)  \qquad
        \text{and} \qquad (\{n^{th} \text{ roots
        of unity}\}, \times) \]
        are realisations of this group, and they
        are all isomorphic.
    \item Let $G$ be a group and $g_1, \dots, gk \in
        G$. Then the subgroup of $G$ generated by
        $g_1, \dots, gk$ denoted by $\langle g_1,
        \dots, g_k \rangle$ is the smallest subgroup of
        $G$ containing all the $g_i$. It is the
        intersection of all the subgroups of $G$
        containing all the $g_i$.
\end{enumerate}

\newpage
\section{The Dihedral and Symmetric Groups}

First note composition of functions is
associative:
\[ f, g, h : X \to X, \qquad x \in X \]
Then
\begin{align*}
    (f \circ (g \circ h))(x) 
    &= f((g \circ h)(x)) \\
    &= f(g(h(x))) \\
    &= (f \circ g)(h(x)) \\
    &= ((f( \circ g) \circ h)(x)
    \implies f \circ (g \circ h)
    &= (f \circ g) \circ h
\end{align*}

\subsection{Dihedral Groups}

Let $P$ be a regular polygon with $n$ sides and
$V$ its set of vertices. We can assume
\[ V = \{e^{2\pi i k/n} : 0 \le k < n\} \]
$n$-th roots of unity in $\CC$. Then the
symmetries of $P$ are the isometries (i.e.
distance preserving maps of $\CC$ that map  $V$ to
$V$. \\\\
We will show that for $n \ge 3$ the set of
symmetries of $P$, under composition form a
nonabelian group of order $2n$. This group is
called the \emph{dihedral group} of order $2n$ and
denoted by $D_{2n}$.

\begin{notation*}
    Sometimes $D_{2n}$ is denoted $D_n$.
\end{notation*}
\noindent
We have already met $D_6$ in example 10. \\\\
Consider $D_8$
\begin{center}
\begin{tsqx}
    ! size(3cm);
    A := dir 135
    B := dir 45
    C := dir -45
    D := dir -135
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $e$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir 45
    B := dir -45
    C := dir -135
    D := dir 135
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $r$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    arc3 origin 0.2*dir(45) 0.2*dir(135) BeginArcArrow
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir -45
    B := dir -135
    C := dir 135
    D := dir 45
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $r^2$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    arc3 origin 0.2*dir(0) 0.2*dir(180) BeginArcArrow
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir -135
    B := dir 135
    C := dir 45
    D := dir -45
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $r^3$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    arc3 origin 0.2*dir(45) 0.2*dir(135) EndArcArrow
\end{tsqx}
\end{center}
\begin{center}
\begin{tsqx}
    ! size(3cm);
    A := dir 45
    B := dir 135
    C := dir -135
    D := dir -45
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $tr^2$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    0.5*(A+B+(0,0.1))--0.5*(C+D-(0,0.1)) dashed
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir -135
    B := dir -45
    C := dir 45
    D := dir 135
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $t$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    0.5*(A+D-(0.1,0))--0.5*(B+C+(0.1,0)) dashed
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir 135
    B := dir -135
    C := dir -45
    D := dir 45
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $tr$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    A--C dashed
\end{tsqx}
\begin{tsqx}
    ! size(3cm);
    A := dir -45
    B := dir 45
    C := dir 135
    D := dir -135
    label $1$ @ A+0.1*dir(A)
    label $2$ @ B+0.1*dir(B)
    label $3$ @ C+0.1*dir(C)
    label $4$ @ D+0.1*dir(D)
    A--B--C--D--A
    label $tr^3$ @ 0.5*(dir(-45)+dir(-135))-(0,0.1)
    B--D dashed
\end{tsqx}
\end{center}
Let $r : P \to P$
\[ z \mapsto e^{2\pi i/n} z\]
\[ t : p \to P \]
\[ z \mapsto \ol{z} \]
These are both isometries.

\begin{align*}
    |r(z) - r(w)|
    &= |e^{2 \pi i/n} z - e^{2 \pi i/n} w| \\
    &= |e^{2\pi i/n}||z - w| \\
    &= |z - w|
\end{align*}

\begin{align*}
    |t(z) - t(w)|^2
    &= |\ol{z} - \ol{w}|^2 \\
    &= (\ol{z} - \ol{w})(z - w) \\
    &= |z - w|^2 \\
    \implies |t(z) - t(w)|
    &= |z - w|
\end{align*}
Note, $r^n = \id = \text{identity}$
\[ \implies r^{-1} = r^{n - 1} \]
and also
\[ t^2 = \id \implies t = t^{-1} \]
\[ tr(z) = e^{-2\pi i/n} \ol{z} = r^{-1}t(z) \]
\[ \implies tr = r^{-1}t \]
We show that the symmetries of $P$ is
\[ \{\underbrace{e = \id, r, r^2, \dots, r^{n -
1}}_{\text{rotations}},
\underbrace{t, rt, \dots, r^{n -
1}t}_{\text{reflections}}\} \]
Then this set under composition of functions gives
the group $D_{2n}$. \\\\
Let $f$ be a symmetry of $P$. Then $f(1) = e^{2\pi
ik/n}$ for some $k$.
\[ \implies r^{-k} \circ f(1) = 1 .\]
So, $g(e^{2\pi i/n}) = e^{2\pi i/n}$ or $e^{-2\pi
i /n}$.
If $g(e^{2\pi i/n} = e^{2\pi i/n}$ then $g$ fixes
1 and $e^{2\pi i/n}$, Also $g$ interchanges
vertices of $P$ so fixes $P$'s centre of mass
\[ \frac{1}{n} = \sum_{k = 0}^{n - 1} e^{2\pi ik/n}
= 0 .\]
So $g$ fixes 0, 1 and $e^{2\pi i/n}$
\[ g = \id \implies f = r^k .\]
If $g(e^{2\pi i/n}) = e^{-2\pi i/n}$ then
\begin{align*}
    t \circ g(e^{2\pi i/n}
    &= e^{2\pi i/n} \\
    t \circ g(1)
    &= 1 \\
    t \circ g (0)
    &= 0 \\
    \implies t \circ g
    &= \id \\
    t \circ r^{-k} \circ f
    &= \id \\
    \implies f
    &= r^k \circ t^{-1} \\
    &= r^k \circ t
\end{align*}
Algebraically we write,
\[ D_{2n} = \langle \underbrace{r,
t}_{\text{generators}} \mid \underbrace{r^n = e, t^2 = e,
trt = r^{-1}}_{\text{relations}} \rangle \]
Finally, $D_2 \cong C_2$ and $D_4 \cong \ZZ/2\ZZ
\times \ZZ/2\ZZ$ are the only abelian dihedral
groups. Also note that $D_\infty$ exists.

\subsection{Symmetric Groups}

Let $X$ be a set. A bijection
\[ f : X \to X \]
is called a \emph{permutation} of $X$. Let
$\mathrm{Sym}(X)$ denote the set of all
permutations of $X$.

\begin{proposition}
    $\mathrm{Sym}(X)$ is a group under composition
    of functions. It is called the symmetric group
    on $X$.
\end{proposition}

\begin{proof}
    \phantom{.}
    \begin{itemize}
        \item Closure - follows from a lemma in
            Numbers \& Sets
        \item identity, define $c(x) = x \,\,\,\,
            \forall x \in X$
        \item Let $f \in \mathrm{Sym}(X)$. As $f$
            is a bijection, $f^{-1}$ exists and is
            a bijection and satisfies
            \[ f \circ f^{-1} = c = f^{-1} \circ f \]
        \item composition of functions is
            associative as shown earlier
    \end{itemize}
\end{proof}

\begin{notation*}[Symmetric Groups]
    Suppose $X$ is finite and $X = |n|$. Then we
    often take $X$ to be the set $\{1, 2, \dots,
    n\}$ and we write $S_n$ for $\mathrm{Sym}(X)$.
    We call $S_n$ the symmetric group of degree
    $n$.
\end{notation*}
\noindent
We'll use double row notation (for now). \\\\
If $\sigma \in S_n$ write
\[
\sigma =
\begin{pmatrix}
    1 & 2 & \cdots & n \\
    \sigma(1) & \sigma(2) & \cdots & \sigma(n)
\end{pmatrix}
\]
For example
\[
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1
\end{pmatrix}
\in S_3
\]
and
\[
\begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    2 & 3 & 1 & 4 & 5
\end{pmatrix}
\in S_5
\]
Composition:
\[
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1
\end{pmatrix}
\circ
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 1 & 3
\end{pmatrix}
=
\text{``}
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 1 & 3 \\
    3 & 2 & 1
\end{pmatrix}
\text{''}
=
\begin{pmatrix}
    1 & 2 & 3 \\
    3 & 2 & 1
\end{pmatrix}
\]
or
\begin{center}
\begin{tsqx}
    ! size(8cm);
    A .= (-1,1)
    B .= (0,1)
    C .= (1,1)
    D .= (-1,0)
    E .= (0,0)
    F .= (1,0)
    G .= (-1,-1)
    H .= (0,-1)
    I .= (1,-1)
    A--E--I
    B--D--H
    C--F--G
    label $=$ @ (1.3,0)
    A' .= (1.6,0.5)
    B' .= (2.6,0.5)
    C' .= (3.6,0.5)
    D' .= (1.6,-0.5)
    E' .= (2.6,-0.5)
    F' .= (3.6,-0.5)
    A'--F'
    B'--E'
    C'--D'
\end{tsqx}
\end{center}

\subsubsection*{Small $n$}

\[ S_1 = \left\{ 
\begin{pmatrix}
    1 \\
    1
\end{pmatrix}
= \{ c \} \right\} \qquad
\text{trivial group} \]
\[ S_2 = \left\{ 
\begin{pmatrix}
    1 & 2 & \\
    1 & 2
\end{pmatrix}
,
\begin{pmatrix}
    1 & 2 \\
    2 & 1
\end{pmatrix}
\cong (\{\pm 1\}, \times) \cong C_2 \right\} .\]
\[ S_3 = \left\{
\begin{pmatrix}
    1 & 2 & 3 \\
    1 & 2 & 3
\end{pmatrix}
\
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1
\end{pmatrix}
,
\begin{pmatrix}
    1 & 2 & 3 \\
    3 & 1 & 2
\end{pmatrix}
,
\begin{pmatrix}
    1 & 2 & 3 \\
    1 & 3 & 2
\end{pmatrix}
,
\begin{pmatrix}
    1 & 2 & 3 \\
    3 & 2 & 1
\end{pmatrix}
,
\begin{pmatrix}
    1 & 2 & 3 \\
    2 & 1 & 3
\end{pmatrix}
\right\} \cong D_6 \]

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item $|S_n| = n!$ because number of choices
        for $\sigma(1)$ is $n$, number of choices
        for $\sigma(2)$ is $n - 1$\dots
    \item For $n \ge 3$ , $S_n$ is not abelian.
        Consider
        \[
        \begin{pmatrix}
            1 & 2 & 3 & 4 & \cdots & n \\
            2 & 3 & 1 & 4 & \cdots n
        \end{pmatrix}
        ,
        \begin{pmatrix}
            1 & 2 & 3 & 4 & \cdots & n \\
            1 & 3 & 2 & 4 & \cdots & n
        \end{pmatrix}
        \]
    \item $D_{2n}$ naturally embeds in $S_n$. For
        example $D_8 \lesssim S_4$
        \[ r =
        \begin{pmatrix}
            1 & 2 & 3 & 4 \\
            2 & 3 & 4 & 1
        \end{pmatrix}
        \qquad t =
        \begin{pmatrix}
            1 & 2 & 3 & 4 \\
            4 & 3 & 2 & 1
        \end{pmatrix}
        \]
\end{enumerate}

``Double row notation is cumbersome and hides
what's going on. We introduce cycle notation.''

\subsubsection*{New Notation}

\begin{definition}
    Let $a_1, \dots, a_k$ be distinct integers in
    $\{1, \dots, n\}$. Suppose $\sigma \in S_n$
    and
    \[ \sigma(a) = \begin{cases}
        a_{i + 1} & \text{if there exists $i$ such
        that $a_i = a$ (taken modulo $k$).} \\
        a & \text{otherwise}
    \end{cases} \]
    Then $\sigma$ is a \emph{$k$-angle} and we
    write $\sigma = (a_1, a_2, \dots, a_k)$. For
    example
    \[ \sigma = (1, 2, 3) =
    \begin{pmatrix}
        1 & 2 & 3 \\
        2 & 3 & 1
    \end{pmatrix}
    \]
\end{definition}

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item 
        \[ (a_1, a_2, \dots, a_k)
        = (a_k, a_1, a_2, \dots, a_{k - 1}) =
        \cdots \]
        We usually write the smallest $a_i$ first.
    \item
        \[ (a_1, a_2, \dots, a_k)^{-1}
        = (a_1, a_k, a_{k - 1}, \dots, a_2 \]
    \item $o(\sigma) = k$, $\sigma$ is like
        rotations of $k$ points
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            a_k ;= dir 120
            a_1 ;= dir 90
            a_2 ;= dir 60
            arc3 origin dir(30) dir(150)
            arc3 origin dir(150) dir(30) dashed
            label $\sigma$ @ (1.5,0)
            //label $\cdot$ @ 1.11*dir(50)
            //label $\cdot$ @ 1.11*dir(40)
            //label $\cdot$ @ 1.11*dir(30)
            //label $\cdot$ @ 1.11*dir(130)
            //label $\cdot$ @ 1.11*dir(140)
            //label $\cdot$ @ 1.11*dir(150)
            arc3 origin 1.3*dir(-30) 1.3*dir(30) BeginArcArrow
        \end{tsqx}
        \end{center}
    \item a 2-cycle is called a
        \emph{transposition}.
\end{enumerate}

\begin{definition}
    Two cycles $\sigma(a_1, \dots, a_k)$ and $\tau
    = (b_1, \dots, b_l)$ are \emph{disjoint} if
    $\{a_1, \dots, a_k\} \cap \{b_1, \dots, b_l\}
    = \emptyset$.
\end{definition}

\begin{lemma}
    If $\sigma, \tau \in S_n$ are disjoint then
    \[ \sigma\tau = \tau\sigma \qquad
    (\sigma \circ \tau = \tau \circ \sigma) .\]
\end{lemma}

\begin{proof}
    If $x \in \{1, \dots, n\} \setminus \{a_1,
    \dots, a_k\} \cup \{b_1, \dots, b_l\}$, then
    \[ (\sigma \circ \tau)(x) = \sigma(\tau(x)) =
    x = (\tau \circ \sigma)(x) .\]
    For $1 \le i \le k - 1$ we have
    \begin{align*}
        (\sigma \circ \tau)(a_i)
        &= \sigma(\tau(a_i)) \\
        &= \sigma(a_i) \\
        &= a_{i + 1}
    \end{align*}
    \begin{align*}
        (\tau \circ \sigma)(a_i)
        &= \tau(\sigma(a_i)) \\
        &= \tau(a_{i + 1}) = a_{i + 1}
    \end{align*}
    And $\sigma \circ \tau(a_k) = a_1$ and $a\tau
    \circ \sigma(a_k) = a_1$. The same argument
    works for the $b_i$. Thus $\sigma \circ \tau$
    and $\tau \circ \sigma$ agree everywhere which
    implies that $\sigma \circ \tau = \tau \circ
    \sigma$.
\end{proof}

\begin{example*}
    \[ (1\ 2)(3\ 4\ 5) = (3\ 4\ 5)(1\ 2) \]
\end{example*}
\noindent
However this is not necessarily true if two cycles
are disjoint.

\begin{example*}
    Consider $\sigma = (1\ 2\ 3)$ and $\tau = (2\
    4)$. Then we have
    \begin{align*}
        \sigma \circ \tau(1) &= \sigma(1) = 2 \\
        \sigma \circ \tau(2) &= \sigma(4) = 4 \\
        \sigma \circ \tau(3) &= \sigma(2) = 1 \\
        \sigma \circ \tau(4) &= \sigma(3) = 3
    \end{align*}
    Hence $\sigma \circ \tau = (1\ 2\ 4\ 3)$ but
    $\tau \circ \sigma = (1\ 4\ 2\ 3)$.
\end{example*}

\begin{example*}
    \phantom{.}\\[-2\baselineskip]
    \[ (1\ 2\ 3)(2\ 3) = (1\ 2)(3) = (1\ 2) \]
    but
    \[ (2\ 3)(1\ 2\ 3) = (1\ 3) \]
\end{example*}

\begin{notation*}
    When using cycle notation, we often suppress
    1-cycles.
\end{notation*}

\begin{theorem}
    Every permutation can be written as a product
    of disjoint cycles (in an essentially unique
    way).
\end{theorem}

\begin{example*}
    \begin{align*}
        \sigma
        &=
        \begin{pmatrix}
            1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
            2 & 4 & 5 & 7 & 6 & 3 & 1 & 9 & 8
        \end{pmatrix} \\
        &= (1\ 2\ 4\ 7)(3\ 5\ 6)(8\ 9)
    \end{align*}
\end{example*}

\begin{proof}
    Let $a_1 \in \{1, 2, \dots, n\} = X$. Consider
    $a_1$, $\sigma(a_1)$, $\sigma^2(a_1)$, \dots.
    Since $X$ is finite there exists a minimal $j$
    such that $\sigma^j(a_1) \in \{a_1,
    \sigma(a_1), \dots, \sigma^{j - 1}(a_1)\}$. We
    claim: $\sigma^j(a_1) = a_1$ since if not we
    can assume
    \[ \sigma^j(a_1) = \sigma^i(a_i) \]
    where $j > i \ge 1$. Then this implies
    \[ \sigma^{j - i}(a_1) = a_1 \]
    which contradicts the minimality of $j$. So,
    $(a_1, \sigma(a_1), \dots, \sigma^{j -
    1}(a_1))$ is a cycle in $\sigma$. If there
    exists $b \in X \setminus \{a_1, \sigma(a_1),
    \dots, \sigma^{j - 1}(a_1)\}$ consider $b,
    \sigma(b), \dots$. Now we can note that
    $(b, \sigma(b),
    \dots, \sigma^{k - 1}(b))$ is disjoint from
    $(a_1, \sigma(a_1), \dots, \sigma^{j -
    1}(a_1))$ since $\sigma$ is a bijection.
    Continue in this way until all elements of $X$
    are reached.
\end{proof}

\begin{lemma}
    Let $\sigma$, $\tau$ be disjoint cycles in
    $S_n$. Then
    \[ o(\sigma \tau) = \mathrm{lcm}\{o(\sigma), o(\tau)\}
    .\]
\end{lemma}

\begin{proof}
    Let $\mathrm{lcm}\{o(\sigma), o(\tau)\}$ so
    $o(\sigma) \mid k$ and $o(\tau) \mid k$. Then
    \begin{align*}
        (\sigma\tau)^k 
        &= \sigma\tau\sigma\tau
        \cdots \sigma\tau \\
        &= \sigma^k\tau^k \\
        &= ee \\
        &= e
    \end{align*}
    \[ \implies o(\sigma\tau) \mid k \]
    Now suppose $o(\sigma\tau) = n$. Then
    \begin{align*}
        (\sigma\tau)^n
        &= e \\
        \implies \sigma^n\tau^n \\
        &= e
    \end{align*}
    But $\sigma$, $\tau$ move different elements
    of $X$ which implies that we must have
    $\sigma^n = e$ and $\sigma^n = e$, which
    implies that $o(\sigma) \mid n$ and $o(\tau)
    \mid n$ which implies that $k \mid n$, and
    hence
    \[ o(\sigma\tau) = \mathrm{lcm}\{o(\sigma),
    o(\tau)\} \]
    as desired.
\end{proof}

\begin{proposition}
    Any $\sigma \in S_n$ (with $n \ge 2$) can be
    written as a product of transpositions.
\end{proposition}

\begin{proof}
    By the previous theorem it is sufficient to
    show that a $k$-cycle can be written as a
    product of transpositions. We can do this
    directly:
    \[ (a_1,\ a_2,\ \dots,\ a_k)
    = (a_1,\ a_2)(a_2,\ a_3) \cdots (a_{k -2},\ 
    a_{k - 1})(a_{k - 1},\ a_1) \]
\end{proof}

\begin{example*}
    \[ (1\ 2\ 3\ 4\ 5) = (1\ 2)(2\ 3)(3\ 4)(4\ 5)
    = (1\ 2)(1\ 2)(1\ 2)(2\ 3)(3\ 4)(4\ 5)
    = (1\ 5)(1\ 4)(1\ 3)(1\ 2) .\]
\end{example*}

Note that the representation as a product of
transpositions is not unique.

\begin{definition}
    Let $\sigma \in S_n$ with $(n \ge 2)$. Then
    the \emph{sign} of $\sigma$, written
    $\mathrm{sgn}(\sigma)$ is $(-1)^k$ where $k$
    is the number of transpositions in some
    expression of $\sigma$ as a product of
    transpositions.
\end{definition}

\begin{lemma}
    The function $\mathrm{sgn} : S_n \to \{\pm
    1\}$ defined by $\sigma \mapsto
    \mathrm{sgn}(\sigma)$ is well-defined. i.e. if
    \begin{align*}
        \sigma
        &= \tau_1 \cdots \tau_a \\
        &= \tau_1' \cdots \tau_b'
    \end{align*}
    with $\tau_i$ and $\tau_i'$ transpositions
    then
    \[ (-1)^a = (-1)^b .\]
\end{lemma}

\begin{proof}
    Let $c(\sigma)$ denote the number of cycles in
    a disjoint cycle decomposition of $\sigma$
    including 1-cycles, so $c(\id) = n$. Let
    $\tau$ be a transposition.
    \begin{claim*}
        \begin{align*}
            c(\sigma\tau) = c(\sigma) \pm 1
            \equiv c(\sigma) + 1 \pmod{2}
        \end{align*}
    \end{claim*}
    Let $\tau = (k,\ l)$. 2 cases:
    \begin{enumerate}[(i)]
        \item $k$, $l$ in different cycles of
            $\sigma$:
            \[ (k, a_1, \dots, a_r)(l, b_1, \dots,
            b_s)(k, l)
            = (k, b_1, b_2, \dots, b_s, l, a_1, \dots,
            a_r) \]
            and hence $c(\sigma\tau) = c(\sigma) -
            1$.
        \item when $k$, $l$ in same cycle in
            $\sigma$ we have
            \[ (k, a_1, \dots, a_r, l, b_1, \dots,
            b_s) (k, l)
            = (k, b_1, \dots, b_s)(l, a_1, \dots, a_r) \]
            \[ \implies c(\sigma\tau) = c(\sigma)
            + 1 .\]
    \end{enumerate}
    Now assume
    \begin{align*}
        \sigma
        &= \id \cdot \tau_1 \cdots \tau_a \\
        &= \id \cdot \tau_1' \cdots \tau_a'
    \end{align*}
    Then
    \begin{align*}
        c(\sigma)
        &\equiv n + a \pmod{2} \\
        &\equiv n + b \pmod{2} \\
        \implies a
        &\equiv b \pmod{2} \\
        \implies (-1)^a
        &= (-1)^b \qedhere
    \end{align*}
\end{proof}

\subsubsection*{Aside}

Subgroup lattice of $D_6 = \{e, r, r^2, t, rt,
r^2t\}$:
\begin{center}
\begin{tsqx}
    label $D_6$ @ (0,1)
    label $<r>$ @ (-1.5,0)
    label $<t>$ @ (-0.5,0)
    label $<rt>$ @ (0.5,0)
    label $<r^2t>$ @ (1.5,0)
    label $\{e\}$ @ (0,-1)
    (-1.5,0.1)--(0,0.9)--(-0.5,0.1)
    (1.5,0.1)--(0,0.9)--(0.5,0.1)
    (-1.5,-0.1)--(0,-0.9)--(-0.5,-0.1)
    (1.5,-0.1)--(0,-0.9)--(0.5,-0.1)
\end{tsqx}
\end{center}
So we just connect subgroups with a line if one is
a subgroup of another.

\begin{theorem}
    Let $n \ge 2$. The map
    \[ \mathrm{sgn} : (S_n, \circ) \to (\{\pm 1\},
    \times) \qquad\qquad \sigma \mapsto
    \mathrm{sgn}(\sigma) \]
    is a well-defined non-trivial homomorphism.
\end{theorem}

\begin{proof}
    \begin{itemize}
        \item Well-defined as proven earlier.
        \item $\mathrm{sgn}\left( (1\ 2) \right) =
            -1$, so non-trivial.
        \item Now we prove that it is a
            homomorphism: \\
            Let $\alpha, \beta \in S_n$ with
            $\mathrm{sgn}(\alpha) = (-1)^k$,
            $\mathrm{sgn}(\beta) = (-1)^k$, so
            there exists transpositions $\tau_i$
            and $\tau_i'$ such that
            \[ \alpha = \tau_1 \cdots \tau_k
            \qquad \beta = \tau_1' \cdots
            \tau_l' \]
            \[ \implies \alpha\beta = \tau_1
            \cdots \tau_k \tau_1' \cdots \tau_l' \]
            \begin{align*}
                \implies \mathrm{sgn}(\alpha\beta)
                &= (-1)^{k + l} \\
                &= (-1)^k (-1)^l \\
                &= \mathrm{sgn}(\alpha)
                \mathrm{sgn}(\beta)
            \end{align*}
    \end{itemize}
\end{proof}

\begin{definition}
    $\sigma$ is an \emph{even} permutation if
    $\mathrm{sgn}(\sigma) = 1$ and an odd
    permutation if $\mathrm{sgn}(\sigma) = -1$.
\end{definition}

\begin{corollary}
    The even permutations of $S_n$ ($n \ge 2$)
    form a subgroup called the \emph{alternating
    group} and denoted $A_n$.
\end{corollary}

\begin{proof}
    \begin{itemize}
        \item Identity: $\id = (1\ 2)(1\ 2) \in
            A_n$.
        \eqitem \begin{align*}
            \mathrm{sgn}(\sigma) = 1
            &= \mathrm{sgn}(\rho) \\
            \implies \mathrm{sgn}(\sigma\rho)
            &=
            \mathrm{sgn}(\sigma)\mathrm{sgn}(\rho)
            = 1
        \end{align*}
            by the previous theorem
        \item If
            \[ \sigma = \tau_1 \cdots \tau_k \]
            then
            \[ \sigma^{-1} = \tau_k \cdots \tau_1 \]
            \[ \implies \mathrm{sgn}(\sigma) =
            \mathrm{sgn}(\sigma^{-1}) \]
        \item Associativity is inherited.
    \end{itemize}
\end{proof}

\begin{example*}
    \phantom{.}\\[-2.5\baselineskip]
    \begin{align*}
        A_4 = \{
        &e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\
        4)(2\ 3), \\
        &(1\ 2\ 3), (1\ 3\ 2), (1\ 2\ 4), (1\
        4\ 2), \\
        &(2\ 3\ 4), (2\ 4\ 3), (1\ 3\ 4), (1\
        4\ 3)\}
    \end{align*}
\end{example*}

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item $|A_n| = \frac{|S_n|}{2} = \frac{n!}{2}$
        (exercise - see later)
    \item cycles of even length are odd, and
        cycles of odd length are even.
    \item $A_n = \mathrm{Ker}(\mathrm{sgn})$, hence a
        subgroup. (question 9, sheet 1)
\end{enumerate}

\newpage
\section{Cosets and Lagrange}

\begin{definition}[Cosets]
    Let $H \le G$ and $g \in G$. The \emph{left
    coset $gH$} is defined to be
    \[ \{gh : h \in H \} .\]
    Similarly the right coset is given by
    \[ Hg = \{ hg : h \in H \} .\]
\end{definition}

\begin{example*}
    \[ S_r = \{e, (1\ 2\ 3), (1\ 3\ 2), (1\ 2),
    (1\ 3), (2\ 3) \} .\]
    \[ H = \{\id, (1\ 2\ 3), (1\ 3\ 2)\} = A_3 .\]
    \[ (1\ 2)H = \{(1\ 2), (1\ 2)(1\ 2\ 3), (1\
    2)(1\ 3\ 2)\} = \{(1\ 2), (2\ 3), (1\ 3)\} \]
    \[ (1\ 2\ 3)H = H \]
    Note, $H \dot\cup (1\ 2)H = S_3$.
\end{example*}

\begin{notation*}
    We sometimes use $\dot\cup$ instead of $\cup$
    if we wish to emphasise that we have a
    disjoint union.
\end{notation*}

\begin{lemma}
    Let $H \le G$ and $g \in G$. Then there is a
    bijection between $H$ and $gH$. In particular
    if $H$ is finite then
    \[ |H| = |gH| .\]
\end{lemma}

\begin{proof}
    Define
    \[ \theta_g : H \to gH \qquad h \mapsto gh \]
    We show $\theta_g$ is a bijection.
    \begin{enumerate}
        \item[surj:] If $gh \in gH$ then
            $\theta_g(h) = gh$.
        \item[inj:] If \begin{align*}
            \theta_g(h_1)
            &= \theta_g(h_2) \\
            \implies gh_1
            &= gh_2 \\
            \implies h_1
            &= h_2
        \end{align*}
    \end{enumerate}
    \begin{lemma}
        The left cosets of $H$ in $G$ form a
        partition called of $G$ i.e.
        \begin{enumerate}[(i)]
            \item each $g \in G$ lies in some left
                coset of $H$ in $G$.
            \item if $aH \cap bH \neq \emptyset$
                for some $a, b \in G$ then $aH =
                bH$.
        \end{enumerate}
    \end{lemma}
    \begin{center}
    \begin{tsqx}
        (-3,-1)--(-3,1)--(3,1)--(3,-1)--(-3,-1)
        (-2,-1)--(-2,1)
        (-1,-1)--(-1,1)
        (0,-1)--(0,1)
        (2,-1)--(2,1)
        label $H = eH$ @ (-2.5,0)
        label $g_1H$ @ (-1.5,0)
        label $g_2H$ @ (-0.5,0)
        label $g_{k - 1}H$ @ (2.5,0)
        label $\cdots$ @ (1,0)
        label $G$ @ (3.1,-1.1)
    \end{tsqx}
    \end{center}
    \begin{proof}
        \begin{enumerate}[(i)]
            \item $g \in gH$.
            \item Suppose $c \in aH \cup bH$. Then
                we claim that $aH = cH = bH$. Now
                $c \in aH$ so $c = ak$ for some $k
                \in H$
                \begin{align*}
                    \implies cH
                    &= \{ch : h \in H\} \\
                    &= \{akh : h \in H\} \subseteq
                    aH
                \end{align*}
                Similarly, $a = ck^{-1} \in cH$
                \[ \implies aH \subseteq cH \]
                So $aH = cH$. Similarly $cH = bH$.
        \end{enumerate}
        For example $S_n = A_n \dot\cup (1\ 2)
        A_n$.
    \end{proof}
    \begin{lemma}
        Let $H \le G$, $a, b \in G$. Then
        \[ aH = bH \iff a^{-1}b \in H .\]
    \end{lemma}
        \begin{enumerate}
            \item[$(\Rightarrow)$] $b \in bH = aH$
                \[ \implies b = ah \qquad\text{for
                some $h \in H$}\]
                \[ \implies a^{-1}b = h \in H \]
            \item[$(\Leftarrow)$] Suppose
                $a^{-1}b = k \in H$.
                \[ \implies b = ak \in aH \]
                also $b \in bH$,
                \[ \implies aH = bH \qquad\text{by
                earlier lemma} \]
        \end{enumerate}
\end{proof}

\begin{theorem}[Lagrange's Theorem]
    Let $H$ be a subgroup of the finite group $G$.
    Then the order of $H$ divides the order of $G$
    (i.e. $|H| \,\big|\, |G|$).
\end{theorem}

\begin{proof}
    By Lemma 10 $G$ is partitioned into distinct
    cosets of $H$, say
    \[ G = g_1H \dot\cup g_2H \dot\cup \cdots
    \dot\cup g_kH \]
    ($g_1 = e$ say) \\
    By Lemma 9
    \[ |g_iH| = |H| \qquad 1 \le i \le k\]
    \[ \implies |G| = |H|k \]
    so the order of $H$ divides the order of $G$.
\end{proof}

\begin{definition}[14]
    Let $H \le G$. The \emph{index} of $H$ in $G$
    is the number of left cosets of $H$ in $G$,
    denoted $|G:H|$.
\end{definition}

\begin{remark*}
    \begin{enumerate}[(i)]
        \item If $G$ is finite, $|G:H| =
            \frac{|G|}{|H|}$. But can have $|G:H|$
            finite but $G$ and $H$ both infinite.
        \item We write $(G:H)$ for the set of left
            cosets of $H$ in $G$.
    \end{enumerate}
\end{remark*}

\begin{corollary}[Lagrange's Corollary]
    Let $G$ be a finite group and $g$ an element
    of $G$. Then $o(g) \,\big|\, |G|$. In
    particular, $g^{|G|} = e$.
\end{corollary}

\begin{proof}
    Note
    \[ \langle g \rangle = \{e, g, \dots, g^{n - 1}\} \]
    where $o(g) = n$. Then
    \[ o(g) = | \langle g \rangle | \,\big|\, |G| \]
    by Lagrange's Theorem
    \[ \implies g^{|G|} = e .\]
\end{proof}

\begin{corollary}
    If $|G| = p$ for some prime $p$, then $G$ is
    cyclic.
\end{corollary}

\begin{proof}
    Let $e \neq g$. Then
    \[ \{e\} \neq \langle g \rangle \le G \]
    BY Lagrange
    \[ 1 \neq | \langle g \rangle | \,\big|\, |G|
    = p .\]
    \[ \implies | \langle g \rangle | = p = |G| \]
    \[ \implies \langle g \rangle = |G| \]
    i.e. $G$ is cyclic.
\end{proof}

\begin{definition}[Euler Totient Function]
    Let $n \in \NN$ then we define
    \[ \varphi(n) = |\{ 1 \le a \le n :
    (a, n) = 1 \}| \]
    so for example $\varphi(12) = |\{1, 5, 7,
    11\}| = 4$.
\end{definition}

\begin{theorem}[Fermat-Euler Theorem]
    Let $n \in \NN$, $a \in \ZZ$ with $(a, n) =
    1$. Then
    \[ a^{\varphi(n)} \equiv 1 \pmod{n} .\]
\end{theorem}

\noindent
Fermat's Little Theorem is a special case: \\
$P$ prime, $a \in \ZZ$, $(a, p) = 1$, then
\[ a^{p - 1} \equiv 1 \pmod{p} .\]

We prove Fermat-Euler Theorem by using Lagrange,
first we need to set it up. Let $n \in \NN$,
\[ R_n = \{0, 1, \dots, n - 1\} \]
\[ R_n^* = \{a \in R_n : (a, n) = 1 \} .\]
Define $\times_n$ to be multiplication modulo $n$.

\begin{claim*}
    $(R_n^*, \times_n)$ is a group.
\end{claim*}

\noindent
Notation, $u \in \ZZ$ then $\ul{u} \in R_n$ such
that $u \equiv \ul{u} \pmod{n}$. Closure:
\[ (a, n) = 1 = (b, n) \implies (ab, n) = 1
\implies (\ul{ab}, n) = 1 \]
Identity is 1, and clearly associative. \\
Inverses: Let $a \in R_n^*$ with $(a, n) = 1$.
\[ \implies \exists u, v \in \ZZ \]
such that $au + vn = 1$ (Bezout's Theorem)
\[ \implies au \equiv 1 \pmod{n} \]
Then $\ul{u} \in R_n^*$ is $a^{-1}$.

\bigskip

\noindent
Now we can prove Fermat Euler Theorem:

\begin{proof}
    Note $|R_n^*| = \varphi(n)$
    \[ a \equiv \ul{a} \pmod{n} \qquad \ul{a} \in
    R_n^* \]
    By Corollary 2
    \[ \ul{a}^{\varphi(n)} = \ul{a}^{|R_n^*|} = 1
    \qquad \text{in $R_n^*$} \]
    \[ \implies a^{\varphi(n)} \equiv 1 \pmod{n}
    \qedhere \]
\end{proof}

\newpage
\section{Normal Subgroups, Quotient Groups and Homomorphisms}

Given a group $G$, subgroup $H$ of $G$ and the set
of left cosets of $H$ in $G$, $(G : H)$, we would
like to define a group operation on the cosets,
$\circ$, so that $((G : H), \circ)$ is a group. We
would like
\[ (gH) \circ (kH) = gkH .\]
When does this work?
\[ gHkH = gkHH = gkH \iff kH = Hk \]
This motivates the following definition:

\begin{definition}[15]
    A subgroup $K$ of $G$ is called \emph{normal}
    if $gK = Kg$ for all $g \in G$. We write $K
    \normalsub G$.
\end{definition}

\begin{example*}
    \[ K = \{\id, (1\ 2\ 3), (1\ 3\ 2)\}
    \normalsub S_3 .\]
    \[ (1\ 2)K = \{(1\ 2), (2\ 3), (1\ 3)\} = K(1\
    2) \]
    \[ (1\ 3)K = K(1\ 3) \]
    \[ (2\ 3)K = K(2\ 3) \]
    And $(1\ 2\ 3)K = K = K(1\ 2\ 3)$ etc. But $H
    = \{1, (1\ 2)\}$ is not normal in $S_3$:
    \[ (1\ 3)H = \{(1\ 3), (1\ 2\ 3)\} \]
    \[ H(1\ 3) = \{(1\ 2), (1\ 3\ 2)\} .\]
\end{example*}

\begin{proposition}[4]
    Let $K \le G$. TFAE (the following are
    equivalent):
    \begin{enumerate}[(i)]
        \item $gK = Kg\,\,\forall g \in G$
        \item $gKg^{-1} = K\,\,\forall g \in G$
        \item $gkg^{-1} \in K\,\,\forall k \in K,
            g \in G$.
    \end{enumerate}
\end{proposition}

\begin{proof}
    (i) $\implies$ (ii):
    \begin{align*}
        gKg^{-1}
        &= \{gkg^{-1} : k \in K\} \\
        &= (gK)g^{-1} \\
        &= (Kg)g^{-1} \\
        &= K
    \end{align*}
    (ii) $\implies$ (iii): trivial. \\
    (iii) $\implies$ (i): For any $k \in K$, $g
    \in G$, there exists $k' \in K$ such that
    \[ gkg^{-1} = k' \]
    \[ \implies gk = k'g \in Kg \]
    \[ \implies gK \subseteq Kg \]
    Similarly $g^{-1}kg = k''$ for some $k'' \in
    K$
    \[ \implies kg = gk'' \]
    \[ \implies Kg \subseteq gK \]
    \[ \implies gK = Kg .\]
\end{proof}

\subsubsection*{Examples}

\begin{itemize}
    \item $\{e\} \normalsub G$, $G \normalsub G$.
    \item If $G$ is abelian then all subgroups are
        normal. Since if $k \in K$, $g \in G$, $K
        \normalsub G$ follows from
        \[ gkg^{-1} = gg^{-1}k = k \in K .\]
    \item Kernels of homomorphisms are normal
        subgroups (Sheet 1, question 9).
        \[ \implies A_n \normalsub S_n \]
        since $A_n = Ker(\mathrm{sgn})$.
    \item $D_{2n} = \langle r, y : r^n = 1 = t^2,
        trt = r^{-1} \rangle$
        Then $\langle r \rangle \normalsub
        D_{2n}$. Clearly $r^ir^jr^{-i} = r^j \in
        \langle r \rangle$. Also
        \begin{align*}
            (r^it)r^j(r^it)^{-1}
            &= r^itr^jtr^{-1} \\
            &= r^i - j - i = r^{-j} \in \langle r
            \rangle
        \end{align*}
        Or we can use the following lemma.
        \begin{lemma}
            If $K \le G$ and the index of $K$ in
            $G$ is 2, then $K \normalsub G$.
        \end{lemma}
        \begin{proof}
            \begin{align*}
                G
                &= K \dot\cup gK \\
                &= K \dot\cup Kg \\
                \implies gK
                &= Kg \,\,\forall g \in G
            \end{align*}
        \end{proof}
\end{itemize}

\begin{theorem}
    If $K \normalsub G$, the set $(G : K)$ of left
    cosets of $K$ in $G$ is a group under coset
    multiplication, i.e.
    \[ gK \cdot hK = ghK \]
    This group is called the \emph{quotient group}
    (or factor group of $G$ by $K$ and denoted $G
    / K$.
\end{theorem}

\begin{proof}
    We need to check that cost multiplication is
    well-defined, i.e. if
    \[ gK = \hat{g}K \]
    and
    \[ hK = \hat{h}K \]
    then
    \[ ghK = \hat{g}\hat{h}K .\]
    By Lemma 11,
    \[ gK = \hat{g}K \implies \hat{g}^{-1}g \in K \]
    \[ hK = \hat{h}K \implies \hat{h}^{-1}h \in K \]
    Now $\hat{g}^{-1}g \in K$
    \[ \implies h^{-1}\hat{g}^{-1}gh \in K \]
    since $K \normalsub G$.
    \[ \implies \hat{h}^{-1}hh^{-1}\hat{g}^{-1}gh
    \in K \]
    \[ \implies \hat{h}^{-1}\hat{g}^{-1}gh \in K \]
    \[ \implies ghK = \hat{g}\hat{h}K \]
    by Lemma 11. So coset multiplication is
    well-defined. Group axioms now follow easily:
    \begin{itemize}
        \item By construction coset multiplication
            is closed as $ghK \in (G : H)$ $g_1h
            \in G$.
        \item identity given by $eK = K$
        \item $(gK)^{-1} = g^{-1}K$.
        \item associativity holds since it does in
            $G$, to check:
            \begin{align*}
                (gKhK)lK
                &= (gh)lK \\
                &= g(hl)K \\
                &= gk(HklK)
            \end{align*}
    \end{itemize}
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $S_n / A_n = (\{A_n, (1\ 2)A_n\}, \circ)
        \cong C_2$.
    \item $D_8 = \langle a, b : a^4 = 1 = b^2, bab
        = a^{-1} \langle$ Let $K = \{1, a^2\}$.
        \begin{claim*}
            $K \normalsub D_8$.
        \end{claim*}
        \begin{align*}
            (a^i b)a^2(a^ib)^{-1}
            &= a^i ba^2ba^{-i} \\
            &= a^{-2} = a^2 \in K \\
            a^ia^2a^{-1}
            &= a^2 \in K
        \end{align*}
        \[ \frac{|D_8|}{|K|} = 4 = |(D_8 : K)| \]
        4 distinct left cosets:
        \begin{align*}
            K
            &= \{1, a^2\} \\
            aK
            &= \{a, a^3\} \\
            bK
            &= \{b, ba^2\} = \{b, a^2b\} \\
            abK
            &= \{ab, aba^2\} = \{ab, a^3b\}
        \end{align*}
        \begin{center}
            \begin{tabular}{c|cccc}
                $\circ$ & $K$ & $aK$ & $bK$ & $abK$ \\
                \hline
                $K$ & $K$ & $aK$ & $bK$ & $abK$ \\
                $aK$ & $aK$ & $K$ & $abK$ & $bK$ \\
                $bK$ & $bK$ & $abK$ & $K$ & $aK$ \\
                $abK$ & $abK$ & $bK$ & $aK$ & $K$
            \end{tabular}
        \end{center}
        Note: $aKaK = a^2K = K \cong \text{example
        9}$.
    \item Recall the subgroups of $(\ZZ, +)$ are
        precisely the groups $(n\ZZ, +)$ where $n
        \in \NN$,
        \[ n\ZZ = \{nk : k \in \ZZ\} .\]
        Since $(\ZZ, +)$ abelian, all subgroups
        are normal, $n\ZZ \normalsub \ZZ$. Suppose
        $n = 5$, cosets given by,
        \begin{align*}
            5\ZZ
            &= \{5k : k \in \ZZ\} \\
            1 + 5\ZZ
            &= \{1 + 5k : k \in \ZZ\} \\
            2 + 5\ZZ
            &= \{2 + 5k : k \in \ZZ\} \\
            3 + 5\ZZ
            &= \{3 + 5k : k \in \ZZ\} \\
            4 + 5\ZZ
            &= \{4 + 5k : k \in \ZZ\}
        \end{align*}
        \[ (1 + 5\ZZ) + (2 + 5\ZZ) = 3 + 5\ZZ .\]
        \[ (3 + 5\ZZ) + (4 + 5\ZZ) = 7 + 5\ZZ
        = 2 + 5\ZZ .\]
        \[ (\ZZ/5\ZZ, \circ) \cong (\{0, 1, 2, 3,
        4\}, +_5) \]
        \[ n + 5\ZZ \to \ul{n} \qquad\text{such
        that}\qquad n \equiv \ol{n} \pmod{5} \]
        $\ol{n} \in \{0, 1, 2, 3, 4\}$.
        Well-defined map: \\
        if $n + 5\ZZ = m + 5\ZZ$ then
        \begin{align*}
            -m + n
            &\in 5\ZZ \\
            \implies -m + n
            &\equiv 0 \pmod{5} \\
            \implies n
            &\equiv m \pmod{5} \\
            \implies \ol{n}
            &= \ol{m}
        \end{align*}
        homomorphism:
        \begin{align*}
            \theta((n + 5\ZZ) + (m + 5\ZZ))
            &= \theta(n + m + 5\ZZ) \\
            &= \ol{n + m} \\
            &= \ol{n} +_5 \ol{m} \\
            &= \theta(n + 5\ZZ) + \theta(m + 5\ZZ)
        \end{align*}
        In general
        \[ (\ZZ/n\ZZ, \circ) \cong
        (\{0, 1, 2, 3, 4\}, +_n) .\]
\end{enumerate}

\noindent
Recall $\theta : G \to H$ is a homomorphism if
\[ \theta(xy) = \theta(x)\theta(y) \]
\[ \mathrm{Im}(\theta) = \{\theta(g) : g \in G\}
\le H \]
\[ \mathrm{Ker}(\theta) = \{g \in G : \theta g =
e_H\} \normalsub G \]

\begin{theorem}[First Isomorphism Theorem]
    Let $G$, $H$ be groups and $\theta : G \to H$
    be a group homomorphism. Then
    $\mathrm{Im}(\theta) \le H$ and
    $\mathrm{Ker}(\theta) \normalsub G$ and $G /
    \mathrm{Ker}(\theta) \cong
    \mathrm{Im}(\theta)$.
\end{theorem}

\begin{definition}[16]
    A group is called \emph{simple} if its only
    normal subgroups are $\{e\}$ and $G$. For
    example $C_p$ for some prime $p$.
\end{definition}

\begin{definition*}[Injection]
    Suppose $f : A \to B$. Then $f$ is
    \emph{injective} if for any $a_1, a_2 \in A$,
    if $f(a_1) = f(a_2)$ then $a_1 = a_2$. (each
    element of $A$ maps to a different element of
    $B$).
\end{definition*}

\begin{definition*}[Surjection]
    Suppose $f : A \to B$. Then $f$ is
    \emph{surjective} if given $b \in B$, $\exists
    a \in A$ such that $f(a) = b$. (every element
    in $B$ is `hit').
\end{definition*}

\begin{definition}[Bijection]
    A function is \emph{bijective} if it is both
    injective and surjective.
\end{definition}

Now we can prove the first isomorphism theorem.

\begin{proof}
    Need to construct an isomorphism $\theta : G /
    \mathrm{Ker}\theta \to \mathrm{Im}\theta$
    where $gK \mapsto \theta(g)$. Let $K =
    \mathrm{Ker}\theta$; need $\theta$
    well-defined: \\
    Suppose $gK = hK$, then
    \begin{align*}
        h^{-1}g
        &\in K \\
        \implies \theta(h^{-1}g)
        &= e_H \\
        \implies \theta(h)^{-1} \theta(g)
        &= e_H \qquad \text{since $\theta$ is a
        homomorphism} \\
        \implies \theta(g)
        &= \theta(h) \\
        \implies \theta(gK) = \theta(hK)
    \end{align*}
    Need $\theta$ a homomorphism:
    \begin{align*}
        \theta(gKhK)
        &= \theta(ghK) \\
        &= \theta(gh) \\
        &= \theta(g) \theta(h) \qquad \text{since
        $\theta$ is a homomorpism} \\
        &= \theta(gK) \theta(hK)
    \end{align*}
    $\theta$ surjective:
    \[ \theta(g) \in \mathrm{Im} \theta \implies
    \theta(gK) = \theta(g) \]
    $\theta$ injective: \\
    Suppose $\theta(gK) = \theta(hK)$ then
    \begin{align*}
        \theta(g)
        &= \theta(h) \\
        \implies \theta(h)^{-1} \theta(g)
        &= e_H \\
        \theta(h^{-1}g)
        &= e_H \\
        \implies h^{-1}g
        &\in K \\
        \implies gK
        &= hK
    \end{align*}
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item $\mathrm{sgn} : S_n \to (\{\pm 1\},
        \times)$ with $\sigma \mapsto
        \mathrm{sgn}(\sigma)$. Then
        \begin{align*}
            \mathrm{Im}(\mathrm{sgn})
            &= (\{\pm 1\}, \times) \\
            \mathrm{Ker}(\mathrm{sgn})
            &= A_n \\
            \implies S_n / A_n
            &\cong (\{\pm 1\},
            \times) \cong C_2 \\
            \implies |A_n|
            &= |S_n| / 2
        \end{align*}
    \item $\theta : (\RR, +) \to (\CC \setminus
        \{0\}, \times)$ defined by $r \mapsto
        e^{2\pi i r}$. Note, $\theta(r + s) =
        \theta(r)\theta(s)$. Also,
        \[ \mathrm{Im}(\theta) = S' = \{z \in \ZZ:
        |z| = 1\} \qquad \text{unit circle} \]
        \[ \mathrm{Ker}(\theta) = (\ZZ, +)
        \normalsub (\RR, +) \]
        \[ (\RR, +) / (\ZZ, +) \cong S' \]
    \item Recall
        \[ \mathrm{GL}_2(\RR) = \{\text{$2 \times 2$
        matrices, entries in $\RR$, $\det \neq 0$}\} \]
        Then we observe that
        $\det : \mathrm{GL}_2(\RR) \to (\RR
        \setminus \{0\}, \times)$, $M \mapsto
        \det(M)$ is a homomorphism since
        \[ \det(AB) = \det(A) \det(B) .\]
        \[ \mathrm{Im}(\det) = (\RR \setminus
        \{0\}, \times) \]
        since
        \[ \det
        \begin{pmatrix}
            \alpha & 0 & \cdots & 0 \\
            0 & 1 & \cdots & 0 \\
            \vdots & \vdots & \ddots & \vdots & \\
            0 & 0 & \cdots & 1
        \end{pmatrix}
        = \alpha \in \RR \setminus \{0\} .\]
        \begin{align*}
            \mathrm{Ker}(\det)
            &= \mathrm{SL}_2(\RR) \\
            &= \{\text{$2 \times 2$ matrices,
            entries in $\RR$, $\det = 1$.}\} \\
            \implies \mathrm{SL}_2(\RR)
            &\normalsub \mathrm{GL}_2(\RR)
        \end{align*}
        and $\mathrm{GL}_2(\RR) /
        \mathrm{SL}_2(\RR) \cong (\RR \setminus
        \{0\}, \times)$.
    \item $\theta : (\ZZ, +) \to (\{0, 1, \dots, n
        - 1\}, +_n)$ with $n \mapsto \ul{n}$.
        \[ \mathrm{Ker}\theta = n\ZZ \]
\end{enumerate}

\begin{remark*}
    Let $K \normalsub G$. Then $K$ is the kernel
    of the natural surjective homomorphism
    \begin{align*}
        \theta : G
        &\to G / K \\
        g \mapsto gK
    \end{align*}
    Thus homomorphic images of $G$ are equivalent
    to quotients of $G$.
\end{remark*}

\begin{lemma}
    A homomorphism $\theta : G \to H$ is injective
    if and only if $\mathrm{Ker} \theta = \{e_G\}$.
\end{lemma}

\begin{proof}
    \begin{enumerate}
        \item[($\Rightarrow$)] Suppose $\theta(g) =
            e_H = \theta(e_G)$. Injective implies
            that $g = e_G$.
        \item[($\Leftarrow$)] \phantom{.}
            \\[-2\baselineskip]
            \begin{align*}
                \theta(g)
                &= \theta(h) \\
                \implies \theta(h)^{-1} \theta(g)
                &= e_H \\
                \implies \theta(h^{-1}g)
                &= e_H \\
                \implies h^{-1}g
                &\in
                \mathrm{Ker}\,\,\theta
                = \{e_G\} \\
                \implies h^{-1}g
                &= e_G \\
                \implies h = g
            \end{align*}
            Recall, $N \normalsub G$, $g \in G$,
            $n \in N$ implies
            \begin{align*}
                gng^{-1}
                &\in N \\
                gng^{-1}
                &= \hat{n} \qquad \text{for some
                $\hat{n} \in N$} \\
                &= gn = \hat{n}g
            \end{align*}
    \end{enumerate}
\end{proof}

\begin{lemma}
    \begin{enumerate}[(i)]
        \item Let $N \normalsub G$ and $H \le G$.
            Then $NH = \{nh : n \in N, h \in H\}
            \le G$.
        \item Let $N \normalsub G$, $M \normalsub
            G$, then
            \[ NM \normalsub G .\]
    \end{enumerate}
\end{lemma}

\begin{proof}
    \begin{enumerate}[(i)]
        \item closure, $nh$, $\ul{n}\ul{h} \in
            NH$, then
            \[ n\ub{h\ul{n}}_{\hat{n}h}\ul{h} =
            n\hat{n}h \ul{h} \in NH \]
            identity: $\id = e = ee \in NH$ \\
            inverse:
            \begin{align*}
                (nh)^{-1}
                &= h^{-1}n^{-1} \\
                &= \hat{n} h^{-1} \qquad \text{for
                some $\hat{n} \in N$.} \\
                &\in NH
            \end{align*}
        \item check normality
            \[ g(nm) g^{-1} = \ub{gng^{-1}}_{\in
            N}\ub{gmg^{-1}}_{\in M} \in NM \]
    \end{enumerate}
\end{proof}

\newpage
\section{Direct products and Small Groups}

\subsection{Direct Products}

Let $H$ and $K$ be groups. We construct the
(external) \emph{direct product}, $H \times K$, to
be the set
\[ \{(h, k) : h \in H, k \in K\} \]
with operation
\[ (h_1, k_1) * (h_2, k_2) = (h_1 *_H h_2, k_1 *_K
k_2) = (h_1h_2, k_1k_2) \]
i.e. componentwise multiplication. \\
Then $(H \times K, *)$ is a group, which can
verify easily as follows:
\begin{enumerate}
    \item[closure] $H$ group implies $h_1h_2 \in
        H$ and $K$ group implies $k_1k_2 \in K$.
    \item[identity] $(e_H, e_K)$
    \item[inverse] $(h, k)^{-1} = (h^{-1},
        k^{-1})$
    \item[associativity] since group operations in
        both $H$ and $K$ are associative.
\end{enumerate}

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item If $H$, $K$ both finite, then $|H \times
        K| = |H||K|$.
    \item $H \times K$ abelian if and only if
        \begin{align*}
            (h_1, k_1) * (h_2, k_2)
            &= (h_2, k_2) * (h_1, k_1)
            \,\,\,\forall h_1, h_2 \in H, k_1, k_2
            \in K \\
            \iff (h_1h_2, k_1k_2)
            &= (h_2h_1, k_2k_1) \\
            \iff h_1h_2 = h_2h_2 \qquad
            &\text{and} \qquad k_1k_2 = k_2k_1 \\
            \iff \text{$H$ abelian }
            &\text{and} \text{ $K$ abelian}
        \end{align*}
    \item $H \cong \{(h, e_K) : h \in H\} \le H
        \times k$ and $K \cong \{(e_H, k) : k \in
        K\} \le H \times K$.
\end{enumerate}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item 
        \begin{align*}
            C_2 \times C_2
            &= \langle x \rangle \times \langle y
            \rangle \\
            &= \{e, x\} \times \{e, y\}
        \end{align*}
        elements $(e, e), (x, e), (e, y), (x, y)$.
        \begin{center}
            \begin{tabular}{c|cccc}
                $\circ$ & $(e, e)$ & $(x, e)$ & $(e, y)$ & $(x, y)$ \\
                \hline
                $(e, e)$ & $(e, e)$ & $(x, e)$ & $(e, y)$ & $(x, y)$ \\
                $(x, e)$ & $(x, e)$ & $(e, e)$ & $(x, y)$ & $(e, y)$ \\
                $(e, y)$ & $(e, y)$ & $(x, y)$ & $(e, e)$ & $(x, e)$ \\
                $(x, y)$ & $(x, y)$ & $(e, y)$ & $(x, e)$ & $(e, e)$
            \end{tabular}
        \end{center}
        Klein 4-group $\cong$ example 9. Note
        $o((x, e)) = o(e, y)) = o(x, y)) = 2$. So
        $C_2 \times C_2 \not\cong C_4$.
    \item However, $C_2 \times C_3 \cong C_6$.
        (sheet 2, question 10)
\end{enumerate}

\begin{lemma}
    Let $(h, k) \in H \times K$ where $H$, $K$
    groups. Then
    \[ o((h, k)) = \mathrm{lcm}(o(h), o(k)) \]
\end{lemma}

\begin{proof}
    Let $n = o((h, k))$ and $m =
    \mathrm{lcm}(o(h), o(k))$. Then $h^m = e_H$,
    $k^m = e_K$. So $(h, k)^m = (h^m, k^m) = (e_H,
    e_K)$ and hence $n \mid m$ by Lemma 5. Also,
    \begin{align*}
        (e_H, e_K)
        &= (h, k)^n \\
        &+ (h^n, k^n) \\
        \implies o(h) \mid n,
        &o(k) \mid n \\
        \implies m
        &\mid n
    \end{align*}
    Thus we know when $C_m \times C_n \cong
    C_{mn}$ (Sheet 2, q10).
\end{proof}

\bigskip

\noindent
Recognising when a group can be written as a
direct product of subgroups is trickier.

\begin{proposition}[5]
    Let $G$ be a group with subgroups $H$ and $K$,
    then if
    \begin{enumerate}[(i)]
        \item each element of $G$ can be written
            as $hk$ for $h \in H$ and $k \in K$;
        \item $H \cap K = \{e\}$;
        \item $hk = kh \,\,\,\forall h \in H, k
            \in K$,
    \end{enumerate}
    Then $G \cong H \times K$ and we call $G$ the
    (internal) direct product of $H$ and $K$.
\end{proposition}

\begin{proof}
    Let $\theta : H \times K \to G$ defined by
    $(h, k) \mapsto hk$. First we check that
    $\theta$ is a homomorphism:
    \begin{align*}
    \theta((h_1, k_1)(h_2, k_2))
    &= \theta((h_1h_2, k_1k_2)) \\
    &= h_1h_2k_1k_2 \\
    &= h_1k_1h_2k_2 \\
    &= \theta((h_1, k_1))\theta((h_1, k_2))
    \end{align*}
    To check that $\theta$ is injective,
    \begin{align*}
        \theta((h_1, k_1))
        &= \theta((h_2, k_2)) \\
        \implies h_1k_2
        &= h_2k_2 \\
        \implies h^{-1}h_1
        &= k_2k_1^{-1} \in H \cap K = \{e\} \\
        \implies h_1 = h_2 \qquad 
        &\text{and} \qquad k_1 = k_2
    \end{align*}
    so $(h_1, k_1) = (h_2, k_2)$. $\theta$ is
    surjective by (i), so $\theta$ is an
    isomorphism as required.
\end{proof}

\begin{remark*}
    There are alternative equivalent definitions
    of internal direct product. $G$ is the
    internal direct product of subgroups $H$ and
    $K$ if
    \begin{enumerate}[(i)']
        \item $H \normalsub G$, $K \normalsub G$;
        \item $H \cap K = \{e\}$;
        \item $HK = G$.
    \end{enumerate}
\end{remark*}

\noindent
Need to show (i), (ii), (iii) are equivalent to
(i)', (ii)', (iii)'.

\begin{enumerate}
    \item[($\Rightarrow$)] we show $K \normalsub
        G$. Let $k \in K$, $g = h_1k_1 \in G$ by
        (i). Then
        \[ gkg^{-1} = h_1k_1 kk_1^{-1}h^{-1}
        = h_1 \ul{k} h^{-1} = \ul{k} \in K \]
        Similarly $H \normalsub G$.
    \item[($\Leftarrow$)] Need to show (iii). Let
        $h \in H$, $k \in K$ and consider
        \[ h^{-1}\ub{k^{-1}hk}_{\in H} \in H
        \qquad \text{since $H \normalsub G$.} \]
        Similarly, this expression is in $K$, so
        \begin{align*}
            h^{-1}k^{-1}hk
            &\in H \cap K = \{e\} \\
            \implies hk = kh
        \end{align*}
\end{enumerate}

\begin{example*}
    $G = \langle a \rangle \cong C_{15}$. Then
    \begin{align*}
        C_5
        &\cong \langle a^3 \rangle = H \normalsub
        G \\
        C_3
        &\cong \langle a^5 \rangle = K \normalsub
        G \\
        H \cap K
        &= \langle a^3 \rangle \cap
        \langle a^5 \rangle = \{e\} \\
        a^k
        &= (a^3)^{2k}(a^5)^{-k} \in HK \\
        \implies C_{15}
        &\cong C_3 \times C_5 \cong K \times H
    \end{align*}
\end{example*}

\subsection{Small Groups}

Recall $D_{2n}$, the symmetries of a regular
$n$-gon, generated by
\[ r : z \mapsto e^{2i \pi/n} z \]
\[ t : z \mapsto \ul{z} \]
Then the elements of $D_{2n}$ are
\[ \{e, \ub{r, \dots, r^{n - 1}}_{\text{rotations}},
\ub{t, rt, \dots, rt^{n -
1}}_{\text{reflection}}\} \]

\bigskip

\noindent
Now suppose $G$ a group, $n \ge 3$ with $|G| = 2n$,
and $\exists b \in G$ with $o(b) = n$ and $a \in
G$, $o(a) = 2$ and $aba = b^{-1}$. Then $G \cong
D_{2n}$. Note $\langle b \rangle \normalsub G$
since of index 2. Also $a \not\in \langle b
\rangle$, since $ab \neq ba$. So $G = \langle b
\rangle \cup \langle b \rangle a = \{e, b, \dots,
b^{n - 1}, a, ba, \dots, b^{n - 1}a\}$.
Furthermore
\begin{align*}
    ab
    &= b^{-1}a \\
    \implies ab^k
    &= (ab)b^{k - 1} \\
    &= b^{-1}ab^{k - 1} \\
    &= b^{-2}ab^{k - 2} \\
    &= \cdots \\
    &= b^{-k}a
\end{align*}
So, $(b^ka)(b^ka) = b^kb^{-k}aa = e$.
We can check that
\begin{align*}
    \theta : D_{2n}
    &\to G \\
    r
    &\mapsto b \\
    t
    &\mapsto a
\end{align*}
is an isomorphism.
\begin{itemize}
    \item $|G| = 1$, $G = \{e\}$.
    \item $|G| = 2 \implies G \cong C_2$ (by
        Lagrange's Theorem)
    \item $|G| = 3 \implies G \cong C_3$
    \item $|G| = 4$, by Lagrange's Theorem, $1
        \neq g \in G$ then $o(g) \mid 4$. If
        $\exists g \in G$ with $o(g) = 4$ then
        this implies $G \cong C_4$. Suppose not.
        Let $1 \neq a \in G \implies o(a) = 2$.
        Then by sheet 1 q7, $G$ is abelian, so
        $C_2 \cong \langle a \rangle \normalsub
        G$. Now let $b \in G \setminus \langle a \rangle$,
        then $C_2 \cong \langle b \rangle
        \normalsub G$. Also, $\langle a \rangle
        \cap \langle b \rangle = \{e\}$. Now
        consider ab:
        \begin{itemize}
            \item if $ab = e \implies a = b^{-1} =
                b$ \contradiction
            \item if $ab = a \implies b = e$
                \contradiction
            \item if $ab = b \implies a = e$
                \contradiction
        \end{itemize}
        So,
        \begin{align*}
            G
            &= \{e, a, b, ab\} \\
            &= \langle a \rangle \langle b \rangle
            \\
            &\cong \langle a \rangle \times
            \langle b \rangle \\
            &\cong C_2 \times C_2
        \end{align*}
        Two groups of order 4: $C_4$ and $C_2
        \times C_2$, both of which are abelian.
    \item $|G| = 5 \implies G \cong C_5$ by
        Lagrange's Theorem.
    \item $|G| = 6$ then $1 \neq g \in G \implies
        o(g) \in \{2, 3, 6\}$ by Lagrange. If all
        non-identity elements have order 2 then
        $|G|$ is a 2-power, \contradiction. So
        there exists $b \in G$ such that $o(b) =
        3$ (Note if $o(g) = 6$ then $o(g^2) = 3$).
        Therefore $C_3 \cong \langle b \rangle
        \normalsub G$ since of index 2. Let $a
        \in G \setminus \langle b \rangle$. Hence
        $a^2 \in \langle b \rangle$. (Consider $a
        \langle b \rangle \in G / \langle b
        \rangle$). If $a^2 = b$ or $b^2$ then
        $o(a) = 6 \implies G \cong C_6$. Now
        suppose $a^2 = e$. Also $aba^{-1} \in
        \langle b \rangle$. If $aba^{-1} = e$ then
        $b = e$ which is a contradiction. If
        $aba^{-1} = b$ then $ab = ba \implies
        o(ab) = 6 \implies G \cong C_2$. If
        $aba^{-1} = b^2$, then in other words we
        have $aba^{-1} = b^{-1}$, so $G = \langle
        a, b : a^2 = b^3 = e, aba^{-1} = b^{-1}
        \rangle \cong D_6$. So there are two
        groups of order 6, they are $C_6$ and $D_6
        \cong S_3$. Note $C_6 \not\cong D_6$ as
        $C_6$ is abelian and $D_6$ is not.
    \item $|G| = 7 \implies G \cong C_7$.
    \item $|G| = 8$. By Lagrange, if $1 \neq g \in
        G$ then $o(g) \in \{2, 4, 8\}$. If all
        non-identity elements have order 2 and
        hence $G$ is abelian. Let $1 \neq a \in
        G$, $C_2 \cong \langle a \rangle
        \normalsub G$.
        Choose $b \not\in \langle a \rangle$,
        \begin{align*}
            \langle a, b\rangle
            &= \{1, a, b, ab\} \\
            &= \langle a \rangle \langle b \rangle
            &\cong \langle a \rangle \times
            \langle b \rangle
        \end{align*}
        Choose $c \in G \setminus \langle a, b
        \rangle$. Then
        \begin{align*}
            G
            &= \langle a, b \rangle \cup \langle
            a, b \rangle c \\
            &= \langle a, b \rangle \langle c
            \rangle \\
            &\cong \langle a, b \rangle \times
            \langle c \rangle \\
            &\cong \langle a \rangle \times
            \langle b \rangle \times \langle c
            \rangle \\
            &\cong C_2 \times C_2 \times C_2
        \end{align*}
        Now suppose $\exists g \in G$ such that
        $o(g) > 2 \implies \exists a \in G, o(a) = 4
        \implies C_4 \cong \langle a \rangle
        \normalsub G$. Let $b \in G \setminus
        \langle c \rangle \implies b^2 \in \langle
        a \rangle$. If $b^2 \in \{a, a^3\}
        \implies o(b) = 8 \implies G \cong C_8$.
        Now, $bab^{-1} \in \langle a \rangle$
        (since $\langle a \rangle G$), so
        $bab^{-1} = a^i$ for some $i$. This
        implies 
        \begin{align*}
            b^2ab^{-2}
            &= ba^ib^{-1} \\
            &= (bab^{-1})^i \\
            &= a^{i^2}
        \end{align*}
        But $b^2 \in \langle a \rangle \implies
        b^2ab^{-2} = a$. Hence $i^2 \equiv 1
        \pmod{4} \implies i \equiv \pm 1
        \pmod{4}$. If $bab^{-1} = a \implies ba =
        ab$ so $G$ is abelian. If $b^2 = e$ then
        \begin{align*}
            G
            &= \langle a \rangle \cup \langle a
            \rangle b \\
            &= \langle a \rangle \langle b \rangle
            \\
            &\cong \langle a \rangle \times
            \langle b \rangle \\
            &\cong C_4 \times C_2
        \end{align*}
        if $b^2 = a^2$ then $(ba^{-1})^2 = e$ then
        \begin{align*}
            G
            &\cong \langle a \rangle \times
            \langle ba^{-1} \rangle \\
            &\cong C_4 \times C_2
        \end{align*}
        Suppose $bab^{-1} = a^{-1}$. Then if $b^2
        = e$ then $G \cong D_8$. However if $b^2 =
        a^2$; we have a new group $Q_8$, the
        quaternion group.
        \begin{definition*}[Quaternion Group]
            $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$
            with $ij = k$, $jk = i$, $ki = j$, $ji
            = -k$, $kj = -i$, $ik = ij$ and $i^2 =
            j^2 = k^2 = -1$. So $o(i) = o(j) =
            o(k) = 4$ and $o(-1) = 2$. Another way
            to define the group is:
            \[ \{ \pm
            \begin{pmatrix}
                1 & 0 \\
                0 & 1
            \end{pmatrix}
            , \pm
            \begin{pmatrix}
                i & 0 \\
                0 & -i
            \end{pmatrix}
            , \pm
            \begin{pmatrix}
                0 & i \\
                i & 0
            \end{pmatrix}
            , \pm
            \begin{pmatrix}
                0 & -1 \\
                1 & 0
            \end{pmatrix}
            \} \le \mathrm{SL}_2(\CC) .\]
            alternatively,
            \[ Q_8 = \langle a, b \mid a^4 = e,
            b^2 = a^2, bab^{-1} = a^{-1} \rangle .\]
        \end{definition*}
        So 5 isomorphism classes of groups of
        order 8:
        \[ \ub{C_8, \quad C_4 \times C_2, \quad C_2
        \times C_2 \times C_2}_{\text{abelian}} \]
        all different, because
        \begin{itemize}
            \item $C_8$ has an element of order 8;
            \item $C_4 \times C_2$ does not have
                an element of order 4;
            \item $C_2 \times C_2 \times C_2$ has
                all elements order 2.
        \end{itemize}
        and $D_8$ and $Q_8$ are non-abelian so
        must be different to these 3. $Q_8$ has 6
        elements of order 4, but $D_8$ only has 2,
        so these are non-isomorphic.
    \item $|G| = 9$. We will show later that
        groups of order $p^2$ with $p$ prime are
        abelian. Either $G \cong C_9$ or all
        non-identity elements have order 3. Choose
        $e \neq a \in G$, $b \in G \setminus
        \langle a \rangle$, then
        \begin{align*}
            G
            &= \langle a \rangle \cup \langle a
            \rangle b \cup \langle a \rangle b^2
            \\
            &= \langle a \rangle \langle b \rangle
            \\
            &\cong \langle a \rangle \times
            \langle b \rangle \\
            &\cong C_3 \times C_3
        \end{align*}
    \item $|G| = 10$, must be either $C_{10}$ or
        $D_{10}$ (question 12, sheet 2)
\end{itemize}

\begin{remark*}
    There are lots and lots of groups of order
    $2^k$; there are about 10 of order 16, and about
    $5 \times 10^{10}$ of order $2^{10}$.
\end{remark*}

\newpage
\section{Group Actions}

It's often easier to understand a group if it's doing
something, permuting elements, rotating a square
etc.

\begin{definition}[16]
    Let $G$ be a group and $X$ a non-empty set. We
    say that $G$ \emph{acts} on $X$ if there is a
    mapping
    \[ \rho : G \times X \to X \qquad (g, x)
    \mapsto \rho(g, x) = g(x) \]
    such that
    \begin{enumerate}[(i)]
        \item[(0)] if $g \in G$, $x \in X$, then
            $\rho(g, x) = g(x) \in X$ (implied by
            notation $\rho : G \times X \to X$)
        \item $\rho(gh, x) = \rho(g, \rho(h,
            x))$ (in shorthand, $gh(x) = g(h(x))$)
        \item $\rho(e, x) = x$ (in
            shorthand, $e(x) = x$)
    \end{enumerate}
\end{definition}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item trivial action $\rho(g, x) = x \forall x
        \in X, g \in G$.
    \item $S_n$ acts on the set $\{1, 2, \dots,
        n\} = X$ by permuting the elements of $X$.
        For example, $S_3$ acts on $\{1, 2, 3\}$:
        \begin{align*}
            \sigma = (1\ 2) \in S_3
            &: \qquad \sigma(1) = 2,
            \quad \sigma(2) = 1,
            \quad \sigma(3) = 3 \\
            \tau = (1\ 3) \in S_3 & \\
            \tau\sigma = (1\ 3)(1\ 2) &= (1\ 2\ 3)
            \\
            (\tau\sigma)(1) = 2 = \tau(2) &=
            \tau(\sigma(1))
        \end{align*}
        Similarly subgroups of $S_n$ act on $X$.
    \item $D_8 = \{e, r, r^2, r^3, t, rt, r^2t,
        r^3t\}$ acts on edges of a square
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            A := (1,1)
            B := (-1,1)
            C := (-1,-1)
            D := (1,-1)
            A--B--C--D--A
            label $a$ @ (0,1.1)
            label $b$ @ (1.1,0)
            label $c$ @ (0,-1.1)
            label $d$ @ (-1.1,0)
            (1.5,0)->>(1.5,0.2)
            (1.5,0)->>(1.5,-0.2)
            label $t$ @ (1.6,0)
            arc3 (1.2,1.2) (1.2,1.2)+0.2*dir(0) (1.2,1.2)+0.2*dir(90) BeginArcArrow
            label $r$ @ (1.2,1.2)
        \end{tsqx}
        \end{center}
        \[ t(a) = c, t(c) = a, t(b) = b, t(d) = d,
        s(a) = b, \dots \]
        Also acts on the vertices of a square
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            A := (1,1)
            B := (-1,1)
            C := (-1,-1)
            D := (1,-1)
            A--B--C--D--A
            label $1$ @ (-1.1,1.1)
            label $2$ @ (1.1,1.1)
            label $3$ @ (1.1,-1.1)
            label $4$ @ (-1.1,-1.1)
            (1.5,0)->>(1.5,0.2)
            (1.5,0)->>(1.5,-0.2)
            label $t$ @ (1.6,0)
            arc3 (1.2,1.2) (1.2,1.2)+0.2*dir(0) (1.2,1.2)+0.2*dir(90) BeginArcArrow
            label $r$ @ (1.2,1.2)
        \end{tsqx}
        \end{center}
        \[ t(1) = 3, t(4) = 1, t(2) = 3, t(3) = 2 \]
    \item $G$ acts on itself by left
        multiplication. This is called the
        \emph{left regular action}.
        \[ G \times \to G \qquad (g, k) \mapsto gk \]
        Check:
        \begin{enumerate}[(i)]
            \item[(0)] $gk \in G$ by closure
            \item $\rho(gh, k) = ghk$,
                $\rho(g, \rho(h, k)) = \rho(g,
                hk) = ghk$. Or, in shorthand
                $(gh)k = ghk$, $g(h(k)) = g(hk) =
                ghk$.
            \item $\rho(e, k) = ek = k$.
        \end{enumerate}
        We also have the \emph{right regular
        action}
        \[ G \times G \to G \qquad (g, k) \mapsto
        kg^{-1} \]
    \item $G$ acts on itself by \emph{conjugation}
        \[ G \times G \to G \qquad \]
        Check:
        \begin{enumerate}[(i)]
            \item[(0)] $gkg^{-1} \in G$
            \item $\rho(gh, k) = (gh)k(gh)^{-1} =
                ghkh^{-1}g{-1}$ and $\rho(g,
                \rho(h, k)) = \rho(g, hkh{-1}) =
                g(hkh^{-1})g^{-1}$
            \item $\rho(e, k) = eke^{-1} = k$.
        \end{enumerate}
    \item Let $N \normalsub G$, then $G$ acts on
        $N$ by conjugation
        \[ G \times N \to N \qquad (g, n) \mapsto
        gng^{-1} \]
        \begin{enumerate}[(i)]
            \item[(0)] $gng^{-1} \in N$ since $N
                \normalsub G$.
            \item as above
            \item as above
        \end{enumerate}
    \item Let $H \le G$, then $G$ acts on the set
        of left cosets, $(G : H)$, of $H$ in $G$.
        Called the \emph{left coset action}
        \[ G \times (G : H) \to (G : H) \qquad (g,
        kH) \mapsto (gkH) \]
        \begin{enumerate}[(i)]
            \item[(0)] $gkH \in (G : H)$
            \item $\rho(gh, kH) = (gh)kH = ghkH$
                and $\rho(g, \rho(h, kH)) =
                \rho(g, hkH) = ghkH$
            \item $\rho(e, kH) = ekH = kH$.
        \end{enumerate}
\end{enumerate}

\begin{remark*}
    Recall a permutation of a set $X$ is a
    bijection of $X$. We have commented that a
    bijection $f : X \to X$ has a 2-sided inverse,
    i.e. there exists $g : X \to X$ such that
    \[ f \circ g(x) = x = g \circ f(x) \quad
    \forall x \in X \]
    Conversely, if $f : X \to X$ is a map with a
    2-sided inverse, then $f$ is a bijection:
    \[ f \circ g(x) = x \quad \forall x \in X \implies
    \text{surjective} \]
    \[ g \circ f(x) = x \quad \forall x \in X
    \implies \text{injective} \]
\end{remark*}

\begin{note*}
    2-sided is necessary, because we can consider
    $\phi : \ZZ \to \ZZ$ defined by $x \mapsto
    2x$ and $\psi \ZZ \to \ZZ$ defined by $2x
    \mapsto x$ and $2x + 1 \mapsto 0$. Then
    $\psi\phi = \id$ but $\phi\psi \neq \id$.
\end{note*}

\begin{lemma}
    Suppose the group $G$ acts on the non-empty
    set $X$. Fix $g \in G$, then $\theta_g : X \to
    X$ defined by $x \mapsto \rho(g, x) = g(x)$ is
    a permutation of $X$, i.e. $\theta_g \in
    \mathrm{Sym}(X)$.
\end{lemma}

\begin{proof}
    Clearly $\theta_g$ is a map from $X$ to $X$.
    We need to show $\theta_g$ is a bijection,
    enough to show it has a 2-sided inverse.
    \begin{align*}
        \theta_{g - 1} \circ \theta_g(x)
        &= \theta_{g - 1}(\rho(g, x)) \\
        &= \rho(g^{-1}(\rho(g, x))) \\
        &= \rho(g^{-1}g, x)
        &&\text{since
        $\rho$ group action} \\
        &= \rho(e, x) \\
        &= x
        &&\forall x \in X
    \end{align*}
    Similarly,
    \[ \theta_g \circ \theta_{g - 1}(c) = x
    \qquad \forall x \in X \]
\end{proof}

\begin{proposition}[6]
    Suppose $G$ acts on the set $X$. Then the map
    \[ \theta : G \to \mathrm{Sym}(X) \qquad g
    \mapsto \theta_g \]
    as in Lemma 16, is a homomorphism.
\end{proposition}

\begin{proof}
    We need to show $\theta$ is a homomorphism,
    i.e. we need
    \[ \theta(gh) = \theta(g) \circ \theta(h) \]
    i.e.
    \[ \theta_{gh} = \theta_g \circ \theta_h .\]
    Let $x \in X$, then
    \begin{align*}
        \theta_{gh}(x)
        &= \rho(gh, x) \\
        &= \rho(g, \rho(h, x)) \\
        &= \theta_g \circ \theta_h(x)
    \end{align*}
    True $\forall x \in X$, so done.
\end{proof}

\begin{remark*}
    Proposition 6 gives us an equivalent
    definition of a group action. If $G$ is a
    group and $X$ a set such that $\theta : g \to
    \mathrm{Sym}(X)$ is a group homomorphism, then
    $\rho : G \times X \to X$ defined by $(g, x)
    \mapsto \theta_g(x)$ where $\theta(g) =
    \theta_g$, is a group action.
\end{remark*}

\begin{remark*}
    Using notation of proposition 6, by first
    Isomorphism Theorem,
    \[ G / \mathrm{Ker} \,\, \theta \cong
    \mathrm{Im}\,\,\theta \le \mathrm{Sym}(X) \]
    Note
    \begin{align*}
        \mathrm{Ker}\,\,\theta
        &= \{g \in G : \theta(g) = \id_X \in
        \mathrm{Sym}(X) \} \\
        &= \{g \in G : \theta_g(x) = \rho(g, x) =
        x \forall x\} \\
        &\normalsub G
    \end{align*}
    i.e. all those elements that fix every element
    of $X$, that act `trivially'. We say the
    action is \emph{faithful} if
    $\mathrm{Ker}\,\,\theta = \{e\}$.
\end{remark*}

\subsubsection*{Examples of Kernels}

\begin{enumerate}[(i)]
    \item Trivial action -
        $\mathrm{Ker}\,\,\theta = G$.
    \item $S_n$ acts on $\{1, \dots, n\}$ -
        faithful
    \item $D_8$ acts on edges - faithful
    \item Left regular action - faithful
    \item Conjugation
        \begin{align*}
            \mathrm{Ker}\,\,\theta
            &= \{g \in G : gkg^{-1} = k
            \forall k \in G\} \\
            &= z(G)
        \end{align*}
        where $z(G)$ is the \emph{centre of
        $G$}. `the elements that commute with
        everything'
    \item conjugation on $N \normalsub G$
        \begin{align*}
            \mathrm{Ker}\,\,\theta
            &= \{g \in G : gng^{-1} = n \forall n
            \in N\} \\
            &= C_G(N)
        \end{align*}
        where $C_G(N)$ is the \emph{centraliser of
        $N$ in $G$}.
    \item Left coset action
        \begin{align*}
            \mathrm{Ker}\,\,\theta
            &= \{g \in G : gkH = kH \forall k \in
            G\} \\
            &= \{g \in G : k^{-1}gk \in H \forall
            k \in G\} \\
            &= \{g \in G : g \in kHk^{-1} \forall
            k \in G\} \\
            &= \bigcap_{k \in G} kHk^{-1} \\
            &= \mathrm{Core}_G(H) \\
            &\normalsub G \\
            &\le H
        \end{align*}
\end{enumerate}

\begin{note*}
    If $\mathrm{Ker}\,\,\theta = \{e\}$ then $G$
    is isomorphic to a subgroup of
    $\mathrm{Sym}(X)$, we write $G \lesssim
    \mathrm{Sym}(X)$. So if $|G|$ does not divide
    $|\mathrm{Sym}(X)|$ then
    $\mathrm{Ker}\,\,\theta \neq \{e\}$.
\end{note*}

\begin{theorem}[Cayley's Theorem]
    Any group $G$ is isomorphic to a subgroup of
    $\mathrm{Sym}(X)$ for some non-empty set $X$.
\end{theorem}

\begin{proof}
    We take $X$ to be $G$ and consider the left
    regular action $G \times G \to G$ defined by
    $(g, h) \mapsto gh$. This is a faithful action
    as $gh = h \,\, \forall h \in G \implies g = e$.
    Thus we have an injective homomorphism
    \[ \theta : G \mapsto \mathrm{Sym}(G) \]
    and $G \lesssim \mathrm{Sym}(G)$ as required.
\end{proof}

\begin{definition}[17]
    Let $G$ act on a set $X$ and $x \in X$. The
    \emph{orbit} of $x \in X$ is given by
    \[ \mathrm{Orb}_G(x) = \{g(x) : g \in G\}
    \subseteq X \]
    i.e. the set of points in $X$ which $x$ can be
    mapped to.
\end{definition}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item trivial action, $\mathrm{Orb}_G(x) =
        \{x\}$.
    \item $S_n$ acts on $\{1, 2, \dots, n\} = X$,
        $\mathrm{Orb}_G(1) = X$. If $H = \langle
        (1\ 2)(3\ 4\ 5) \rangle$ acting on $X =
        \{1, 2, 3, 4, 5\}$ then
        \[ \mathrm{Orb}_G(1) = \{1, 2\} \]
        \[ \mathrm{Orb}_G(3) = \{3, 4, 5\} .\]
    \item $D_8$ on \hspace{2.8em}: \\
        \vspace{-3.3em}

        \hspace{2em}
        \begin{tsqx}
            ! size(1cm);
            (1,1)--(1,-1)--(-1,-1)--(-1,1)--(1,1)
            label $a$ @ (0,1.5)
            label $b$ @ (1.5,0)
            label $c$ @ (0,-1.5)
            label $d$ @ (-1.5,0)
        \end{tsqx}
        \[ \mathrm{Orb}_{D_8}(a) = \{a, b, c, d\}
        .\]
    \item left regular action
        \[ \mathrm{Orb}_G(k) = G \]
        since $g = g(k^{-1}k) = (gk^{-1})k$ for
        any $g \in G$.
    \item conjugation
        \begin{align*}
            \mathrm{Orb}_G(k)
            &= \{g(k) : g \in G\} \\
            &= \{gkg^{-1} : g \in G\} \\
            &= \mathrm{ccl}_G(k)
        \end{align*}
        \emph{conjugacy class of $k$ in $G$}. If
        $h \in \mathrm{ccl}_G(k)$ we say $h$ and
        $k$ are \emph{conjugate}.
\end{enumerate}

\begin{definition}[18]
    We say $G$ acts \emph{transitively} on $X$ if
    for any $x \in X$, $\mathrm{Orb}_G(x) = X$.
    Equivalently, if given any pair $x_1, x_2 \in
    X$ $\exists g \in G$ such that $g(x_1) = x_2$.
\end{definition}

\noindent
So, the left regular action is a transitive
action.

\begin{lemma}
    The distinct $G$-orbits form a partition of
    $X$.
\end{lemma}

\begin{proof}
    Let $x \in X$, then $x \in \mathrm{Orb}_G(x)$
    since $x = ex$. Suppose $z \in
    \mathrm{Orb}_G(x) \cap \mathrm{Orb}_G(y)$, we
    show
    \[ \mathrm{Orb}_G(x) = \mathrm{Orb}_G(z) =
    \mathrm{Orb}_G(y) .\]
    $z \in \mathrm{Orb}_G(x) \implies \exists g
    \in G$ such that $g(x) = z$. Suppose $t \in
    \mathrm{Orb}_G(x)$, then $\exists h \in G$
    such that $h(z) = t$ and hence $t = h(g(x)) =
    (hg)(x)$. Therefore $t \in \mathrm{Orb}_G(x)$
    and hence $\mathrm{Orb}_G(z) \subseteq
    \mathrm{Orb}_G(x)$. Similarly $g(x) = z$
    \[ x = e(x) = (g^{-1}g)(x) = g^{-1}(z) \]
    and hence $\mathrm{Orb}_G(x) \subseteq
    \mathrm{Orb}_G(z)$. Thus $\mathrm{Orb}_G(x) =
    \mathrm{Orb}_G(z)$. Similarly
    $\mathrm{Orb}_G(z) = \mathrm{Orb}_G(y)$.
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item We could have proved Lemma 17 by noting
        that $x_1 \sim x_2$ if $\exists g \in G$
        such that $g(x_1) = x_2$ is an equivalence
        relation.
    \item $\mathrm{Orb}_G(x)$ is $G$ invariant,
        i.e.
        \[ g(\mathrm{Orb}_G(x)) \subseteq
        \mathrm{Orb}_G(x) .\]
        Since if $y \in \mathrm{Orb}_G(x)$, then
        $y = hx$ for some $h \in G$.
        \begin{align*}
            \implies g(y)
            &= g(h(x)) \\
            &= (gh)(x) \in \mathrm{Orb}_G(x)
        \end{align*}
    \item $G$ is transitive on
        $\mathrm{Orb}_G(x)$. Let $y, z \in
        \mathrm{Orb}_G(x)$, so $y = g(x)$, $z =
        h(x)$ for some $g, h \in G$. Then
        \[ z = h(g^{-1}(y)) \]
\end{enumerate}

\begin{definition*}[19]
    Let $G$ act on $X$ and $x \in X$. The
    \emph{stabiliser} of $x$ in $G$ is given by
    \[ \mathrm{Stab}_G(x) = \{g \in G : g(x) = x\}
    \subseteq G .\]
    i.e. all those elements in $G$ that fix $x$.
\end{definition*}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item trivial action,
        \[ \mathrm{Stab}_G(x) = G .\]
    \item $S_n$ on $X = \{1, 2, \dots, n\}$
        \[ \mathrm{Stab}_G(1) \cong S_{n - 1} \]
        $H = \langle (12)(345) \rangle$ on $X$
        \begin{align*}
            \mathrm{Stab}_H(1)
            &= \langle (345) \rangle \\
            &= \{e, (345), (354)\}
        \end{align*}
    \item $D_8$ on edges of a square,
        \[ \mathrm{Stab}_{D_8}(e) = \{e, t\} \]
    \item left regular action
        \[ \mathrm{Stab}_G(k) = \{e\} \]
        \[ gk = k \implies g = e \]
    \item conjugation
        \begin{align*}
            \mathrm{Stab}_G(k)
            &= \{g \in G : g(k) = k\} \\
            &= \{g \in G : gkg^{-1} = k\} \\
            &= \{g \in G : gk = kg\} \\
            &= C_G(k)
        \end{align*}
        \emph{centraliser of $k$ in $G$} i.e. all
        elements of $G$ that commute with $k$.
\end{enumerate}

\begin{lemma}
    $\mathrm{Stab}_G(x)$ is a subgroup of $G$.
\end{lemma}

\begin{proof}
    \begin{itemize}
        \item $e(x) = x \implies e \in
            \mathrm{Stab}_G(x)$
        \item if $g, h \in \mathrm{Stab}_G(x)$
            then
            \begin{align*}
                (gh)(x)
                &= g(h(x)) \\
                &= g(x) \\
                &= x \\
                \implies gh
                &\in \mathrm{Stab}_G(x)
            \end{align*}
        \item $g \in \mathrm{Stab}_G(x)$
            \[ g(x) = x \]
            \[ x = e(x) = (g^{-1}g(X) = g^{-1}(gx)
            = g^{-1}(x) \]
            \[ \implies g^{-1} \in
            \mathrm{Stab}_G(x) \]
        \item associativity inherited from $G$.
    \end{itemize}
\end{proof}

\begin{remark*}
    Recall $\phi : G \to \mathrm{Sym}(x)$
    \begin{align*}
        \mathrm{Ker}\,\,\theta
        &= \{g \in G : g(x) = x \,\,\forall x \in
        X\} \\
        &= \cap \mathrm{Stab}_G(x)
    \end{align*}
\end{remark*}

\begin{theorem}[Orbit-Stabiliser Theorem]
    Let $G$ be a finite group acting on a
    non-empty set $X$. Then $\mathrm{Stab}_g(x)
    \le G$ and
    \[ |G| = |\mathrm{Stab}_G(x)||\mathrm{Orb}(x)|
    .\]
\end{theorem}

\begin{remark*}
    We actually prove that $|G :
    \mathrm{Stab}_G(x)|$, the number of left
    cosets of $\mathrm{Stab}_G(x)$ in $G$, is
    equal to $|\mathrm{Orb}_G(x)|$, a more general
    statement.
\end{remark*}

\begin{proof}
    $(G : \mathrm{Stab}_G(x))$ set of left cosets
    of $\mathrm{Stab}_G(x)$ in $G$. Consider the
    map
    \[ \theta : \mathrm{Orb}_G(x) \to (G :
    \mathrm{Stab}_G(x) \qquad g(x) \mapsto
    g\mathrm{Stab}_G(x) \]
    $\theta$ well-defined because:
    \begin{align*}
        g(x)
        &= h(x)
        \implies h^{-1}g(x)
        &= x \\
        \implies h^{-1}g
        &\in \mathrm{Stab}_G(x) \\
        \implies g\mathrm{Stab}_G(x)
        &= h\mathrm{Stab}_G(x) \\
        \implies \theta(g(x))
        &= \theta(h(x))
    \end{align*}
    $\theta$ injective:
    \begin{align*}
        \theta(g(x))
        &= \theta(h(x)) \\
        \implies g\mathrm{Stab}_G(x)
        &= h\mathrm{Stab}_G(x) \\
        \implies h^{-1}g
        &\in \mathrm{Stab}_G(x) \\
        \implies h^{-1}g(x)
        &= x \\
        \implies g(x)
        &= h(x)
    \end{align*}
    $\theta$ surjective: \\
    Given $g\mathrm{Stab}_G(x) \in (G :
    \mathrm{Stab}_G(x))$ then $g(x) \in
    \mathrm{Orb}_G(x)$ and
    \[ \theta(g(x)) = g\mathrm{Stab}_G(x) .\]
    Thus $\theta$ a well-defined bijection as
    required.
\end{proof}

\subsection{Applications to Symmetry Groups of Regular Solids}

Let $S$ be a regular solid and $V$ its vertices.
Then the symmetries of $S$ are the isometries

(distance preserving maps) of $\RR^2$ or $\RR^3$
that maps $S$ to itself.

\subsubsection*{Examples of Symmetries}

\textbf{Example. }(Tetrahedron) \\
This is self-duel. Let $G$ be group of
symmetries of $T$, and $X = \{\text{vertices
of $T$}\} = \{1, 2, 3, 4\}$.
\begin{center}
\begin{tsqx}
    ! size(8cm);
    label $T =$ @ (-1.2,0)
    A := (-1,-1)
    B := (0,1)
    C := (0.5,-0.5)
    D := (0.8,0.7)
    A--B--C--A--D--C
    B--D
    label 4 @ A+0.1*dir(A)
    label 1 @ B+0.1*dir(B)
    label 3 @ C+0.1*dir(C)
    label 2 @ D+0.1*dir(D)
    label faces are 4 @ (3,0.15)
    label equilateral triangles @ (3,-0.15)
\end{tsqx}
\end{center}
Then $\exists$ group homomorphism
\[ \phi : G \to \mathrm{Sym}(X) \cong S_4 \]
(Proposition 6). Note $\mathrm{Ker}\,\,\phi =
\{e\}$, if all vertices fixed, then
$T$ fixed. \\
Consider $G' \le G$ subgroup of rotations.
\begin{center}
\begin{tsqx}
    ! size(8cm);
    // label $T =$ @ (-1.2,0)
    A := (-1,-1)
    B := (0,1)
    C := (0.5,-0.5)
    D := (0.8,0.7)
    A--B--C--A--D--C
    B--D
    (0,-1.2)--(0,1.3) dashed
    arc3 (0,1.3) (0,1.3)+0.3*dir(60) (0,1.3)+0.3*dir(120) EndArcArrow
    label 4 @ A+0.1*dir(A)
    label 1 @ B+0.1*dir(B)+(-0.1,0)
    label 3 @ C+0.1*dir(C)
    label 2 @ D+0.1*dir(D)
    label rotation of $\frac{2\pi}{3}$ @ (3,0.3)
    label 3-cycle (2 3 4) @ (3,0)
    label and $\frac{4\pi}{3}$ (2 4 3) @ (3,-0.3)
\end{tsqx}
\end{center}
4 such axes implies 8 rotations of order 3
(3-cycles).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/8921051843a211ec.png}
\end{center}
3 such axes and identity
\[ \implies G^+ \cong A_4 \]
Now consider $G$ (all symmetries). Clearly
\begin{align*}
    \mathrm{Orb}_G(1)
    &= \{1, 2, 3, 4\} \\
    &= \mathrm{Orb}_{G^+}
\end{align*}
Consider $\mathrm{Stab}_G(1)$. Note if 3
vertices are fixed then $T$ fixed.
Consider $\mathrm{Stab}_G(1)$. Note if 3
Suppose vertices 1 and 2 are fixed.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0711dc2243a311ec.png}
\end{center}
If just 1 fixed have order 3 rotation from
before $= \sigma$. This is everything
\begin{align*}
    \mathrm{Stab}_G(1)
    &= \langle \sigma, \tau \rangle \\
    &\cong D_6 \\
    \implies |G|
    &= |\mathrm{Orb}_G(1)||\mathrm{Stab}_G(1)| \\
    &= 4 \times 6 \\
    &= 24 \\
    \implies G
    &\cong S_4
\end{align*}
Note $\mathrm{Stab}_{G^+}(1) = \langle G \rangle$.
Also $(1234) = (12)(234)$.

\bigskip

\noindent
\textbf{Example. }(Cube) \\
Dual to octahedron.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7fb8bc6843a311ec.png}
\end{center}
Let $G^+$ be group of rotations of $C$. Then $G^+$
acts on set of diagonals $X = \{D_1, D_2, D_3,
D_4\}$.
If a rotation $\sigma$ fixes all
diagonals, then $\sigma = \id$. So we have an
injective homomorphism
\[ \phi : G^+ \to \mathrm{Sym}(C) \cong S_4 \]
roatations: $-\id$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/ee37a9c443a311ec.png}
\end{center}
3 such axes, hence 6 elements of order 4, 3
elements of order 2.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/86a18bee43a411ec.png}
\end{center}
4 such axes, hence 8 elements of order 3.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/587cf54643a411ec.png}
\end{center}
6 such axes, i.e. $G^+ \cong S_4$. \\
Note
$\mathrm{Orb}_{G^+}(D_1) = \{D_1, D_2, D_3,
D_4\}$
\[ \mathrm{Stab}_{G^+}(D_1) = \langle \sigma,
\tau' \rangle \]
or consider $G^+$ acting on vertex 1
\[ |\mathrm{Orb}_{G^+}(1)| = 8 \]
\[ |\mathrm{Stab}_G(1)| = |\langle \rho \rangle| =
3 \]
\[ \implies |G^+| = 24 \]
Now consider full symmetry group of $C$, call it
$G$. Consider action on faces $F_1, \dots., F_6$.
Yields an injective homomorphism (faithful)
\[ \phi : G \to \mathrm{Sym}\{F_i\} \cong S_6 \]
\[ |\mathrm{Orb}(F_1)| = 6 \]
\[ \mathrm{Stab}(F_1) \cong D_8 \]
\[ \implies |G| = 6 \times 8 = 48 .\]
So, action on diagonals is not faithful;
\[ \exists g \in G \quad g(D_i) = D)i) \qquad i
\le i \le 4 \]
but $g \neq \id$. Label vertices of $C$ as
$\{(\pm 1, \pm 1, \pm 1)\}$
\[ g : (x, y, z) \mapsto (-x, -y, -z) \]
if label faces of cube as a dice; 1 opposite 6, 2
opposite 5, 3 opposite 4 then
\[ g = (16)(25)(34) \]
Then $G \cong F^+ \times \langle g \rangle$. Then
$G^+ \normalsub G$ (index 2) and $\langle g
\rangle \normalsub G$ (commutes with all
rotations) and
\[ G^+ \cap \langle g \rangle = \{e\} \]
\[ |G^+ \langle g \rangle| = 48 = |G| .\]

\bigskip

\noindent
\textbf{Example. }(Dodecahedron) \\
Dual to icosahedron. We denote by $D$.
12 regular pentagonal faces, 30 edges, 20
vertices. Let $G^+$ be the grou pof rotations of
$D$. Let $F$ be a face of $D$.
\[ |\mathrm{Orb}_{G^+}(F)| = 12 \]
\[ |\mathrm{Stab}_{G^+}(F)| = 5 \]
\[ \implies |G^+| = 5 \times 12 = 60 \]
There are five cubes embedded in $D$:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/224b83dc43a611ec.png}
\end{center}
$G^+$ acts faithfully on cubes
\[ \implies \phi : G^+ \to S_5 \]
injective and $|G^+| = 60$ hence $G^+ \cong A_5$
(there is some work in the ``hence'' here but one
can do it with some determination). Can find
elements of $A_5$:
\begin{itemize}
    \item rotations through opposite faces - 5
        cycles. (6 axes, 4 elements per axis)
    \item rotation through opposite vertices - 3
        cycles.
    \item rotation through opposite edges - double
        transpositions (15 such).
\end{itemize}
Another application of the Orbit Stabiliser
Theorem:

\setcounter{customtheorem}{7}
\begin{theorem}[Cauchy's Theorem]
    Let $G$ be a finite group and $p$ a prime that
    divides $|G|$. Then there exists an element
    in $G$ of order $p$.
\end{theorem}

\begin{proof}
    Let
    \[ X = \{(x_1, x_2, \dots, x_p) : x_1, x_2,
    \dots, x_p = e, x_i \in G \} .\]
    Let $H = \langle h : h^p = e \rangle \cong
    C_p$ act on $X$ as follows:
    \[ H \times X \to X \qquad (h, (x_1, \dots,
    x_p)) \mapsto (x_2, x_3, \dots, x_p, x_1) \]
    in general,
    \[ (h^i, (x_1, \dots, x_p)) \mapsto (x_{1 +
    i}, x_{2 + i}, \dots, x_{p + i}) \]
    where suffices are taken modulo $p$. \\
    Check this is a group action:
    \begin{enumerate}[(i)]
        \item[(0)] Since $x_1x_2 \cdots x_p = e$, we
            have
            \begin{align*}
                x_1x_2 \cdots x_p
                &= (x_1x_2 \cdots x_i)^{-1} x_1x_2
                \cdots x_p (x_1x_2 \cdots x_i) \\
                &= (x_1x_2 \cdots x_i)^{-1}e(x_1x_2
                \cdots x_i) \\
                &= e
            \end{align*}
        \item We simply check that
            \begin{align*}
                h^{i + j}
                &= (x_{1 + i + j}, \dots, x_{p + i
                + j}) \\
                &= h^i(h^j(x_1, \dots, x_p))
            \end{align*}
        \item For identity, we heck that
            \begin{align*}
                e(x_1, \dots, x_p)
                &= h^p(x_1, \dots, x_p) \\
                &= (x_1, \dots, x_p)
            \end{align*}
    \end{enumerate}
    Let
    \[ \ol{x} = (x_1, x_2, \dots, x_p) \in X .\]
    As distinct orbits partition $X$ (Lemma 17)
    \[ \implies \sum_{\substack{\text{distinct}\\
    \text{orbits}}} |\mathrm{Orb}_H(\ol{x})| = |X| \]
    Note $|X| = |G|^{p - 1}$ (choose $x_1, \dots,
    x_{p - 1}$ then $x_p$ determined)
    \[ \implies p \,\big|\, |X| \]
    \[ \implies p \,\big|\, LHS \]
    But by Orbit Stabiliser Theorem:
    \begin{align*}
        |\mathrm{Orb}_H(\ol{x})|
        &\,\big|\, |H| = p \\
        \implies |\mathrm{Orb}_H(\ol{x})|
        &= 1 \text{ or } p
    \end{align*}
    Now,
    \[ \ol{e} = (e, e, \dots, e) \in X \qquad
    |\mathrm{Orb}_H(\ol{e})| = 1 .\]
    So there exists at least $p - 1$ other orbits
    of length 1. So there exists $\ol{x} \in X$
    with $\mathrm{Orb}_H(\ol{x}) = 1$
    \[ \implies \ol{X} = (x, x, \dots, x) \]
    so $x \neq e$ and $x^p = e$.
\end{proof}

\subsection{Conjugacy Action}

Reminder of the definition of conjugation:
\[ G \times G \to G \qquad (g, h) \mapsto ghg^{-1} .\]
orbits are called conjugacy classes:
\[ \mathrm{ccl}_G(h) = \{ghg^{-1} : g \in G\} .\]
Stabilisers are called centralisers:
\[ C_G(h) = \{g \in G : ghg^{-1} = h\} .\]

\subsubsection*{Remarks}

\begin{enumerate}[(i)]
    \item By Lemma 17 the conjugacy classes
        partition $G$.
    \item By Orbit Stabiliser Theorem, $h \in G$
        \[ |G| = |C_G(h)| |\mathrm{ccl}_G(h)| .\]
        In particular,
        \[ |\mathrm{ccl}_G| \,\big|\, |G| .\]
    \item If $k \in \mathrm{ccl}_G(h)$ then $o(k)
        = o(h)$. Since $k = ghg^{-1}$ for some $g
        \in G$,
        \begin{align*}
            k^{o(h)}
            &= (ghg^{-1})^{o(h)} \\
            &= gh^{o(h)}g^{-1} \\
            &= e \\
            \implies o(k) \mid o(h)
        \end{align*}
        Similarly, $h = g^{-1}kg$ hence $o(h) \mid
        o(k)$, so $o(h) = o(k)$ as desired.
    \item Recall
        \begin{align*}
            Z(G)
            &= \{g \in G : gh = hg \,\,\forall
            h \in G\} \\
            &\normalsub G
        \end{align*}
        And,
        \[ Z(G) = \bigcap_{h \in G} C_G(h) \]
        Note, $z \in Z(G)$ if and only if
        $|\mathrm{ccl}_G(z)| = 1$. If $z \in Z(G)$
        \[ \implies \mathrm{ccl}_G(z) = \{gzg^{-1}
        : g \in G\} = \{z : g \in G\} = \{z\} .\]
        If $|\mathrm{ccl}_G(z)| = 1$ then note
        \[ z = eze^{-1} \in \mathrm{ccl}_G(z) .\]
        So $gzg^{-1} = z \,\,\forall g \in G$.
    \item Let $H \le G$, then $H$ is normal if and
        only if it is a union of conjugacy
        classes. (Sheet 3 question 3)
    \item $G$ abelian if and only if $G = Z(G)$.
\end{enumerate}

\setcounter{customproposition}{6}
\begin{proposition}
    Let $p$ a prime and $G$ a group of order $p^n$.
    Then $Z(G)$ is nontrivial, i.e. $Z(G) \gneq
    \{e\}$.
\end{proposition}

\begin{proof}
    Let $G$ act on $G$ by conjugation. Then the
    conjugacy classes of $G$ partition it by Lemma
    17:
    \[ G = \bigcup_{\substack{\text{distinct}\\
    \text{conjugacy}\\\text{classes}}}
    \mathrm{ccl}_G(x) \]
    By Orbit Stabiliser Theorem
    \[ |\mathrm{ccl}_G(x)| \,\big|\, |G| = p^n .\]
    Either $|\mathrm{ccl}_G(x) = 1$ or $p
    \,\big|\, \mathrm{ccl}_G(x)$. So by (iv) above
    \[ |G| = \sum_{x \in Z(G)} |\mathrm{ccl}_G(x)|
    + \sum_{\substack{\text{distinct}\\
    \text{conjugacy}\\\text{classes}\\\text{with}\\
    \text{$p \,\big|\, \mathrm{ccl}_G(x)$}}}
    |\mathrm{ccl}_G(x)| \]
    Now $p \mid LHS$ so $p \mid RHS$
    \[ \implies p \,\Bigg|\, \sum_{z \in Z(G)}
    |\mathrm{ccl}_G(x)| = |Z(G)| .\]
    But $e \in Z(G)$, hence we must have $|Z(G)|
    \ge p > 1$, as desired.
\end{proof}

\begin{lemma}
    Let $G$ be a finite group and $Z(G)$ the
    centre of $G$. If $G / Z(G)$ is cyclic then
    $G$ is abelian.
\end{lemma}

\begin{proof}
    Let $Z = Z(G)$. Since $G / Z$ is cyclic, $G /
    Z = \langle yZ \rangle$ for some $y \in G$.
    Let $g, h \in G$. Then $gZ = y^iZ$ for some
    $i$, so $g = z^iz_1$ for some $z_1 \in Z$.
    Similarly, $hZ = y^jZ$ for some $j$, so $g =
    z^jz_2$ for some $z_2 \in Z$. Now,
    \begin{align*}
        gh
        &= y^iz_1y^jz_2 \\
        &= y^iy^jz_1z_2
        &&z_1 \in Z \\
        &= y^jy^iz_2z_1 \\
        &= y^jz_2y^iz_1 \\
        &= hg
    \end{align*}
    so $G$ is abelian as required.
\end{proof}

\setcounter{customcorollary}{4}
\begin{corollary}
    Suppose $|G| = p^2$ for some prime $p$. Then
    $G$ is abelian and there are, up to
    isomorphism, just two groups of order $p^2$,
    namely $C_p \times C_p$ and $C_{p^2}$.
\end{corollary}

\begin{proof}
    (Sheet 3 Question 10)
\end{proof}

\subsubsection*{Remark}

\begin{enumerate}[(i)]
    \item A group of order $p^n$ for a prime $p$
        is called a finite $p$-group.
    \item If all elements have $p$-power order $G$
        is called a $p$-group. For example
        $C_{p^\infty}$ (Pr\"ufer group).
\end{enumerate}

\subsubsection*{Conjugation in $S_n$}

\setcounter{customdefinition}{19}
\begin{definition}
    Let $\sigma \in S_n$ and write $\sigma$ as a
    product of disjoint cycles including 1-cycles.
    Then the \emph{cycle-type} of $\sigma$ is
    $(n_1, n_2, \dots, n_k)$ where $n_1 \ge n_2
    \ge \cdots \ge n_k \ge 1$ and the cycles in
    $\sigma$ have length $n_i$. Note $n = n_1 +
    n_2 + \cdots + n_k$. For example
    \[ (1234)(567) = (1234)(567(8) \in S_8 \]
    has cycle type $(4, 3, 1)$, and $e \in S_5$
    has cycle type $(1, 1, 1, 1, 1)$.
\end{definition}

\setcounter{customtheorem}{8}
\begin{theorem}
    The permutations $\pi$ and $\sigma$ in $S_n$
    are conjugate \emph{in $S_n$} if and only if
    they have the same cycle type.
\end{theorem}

\begin{proof}
    Suppose $\sigma$ has cycle type $(n_1, n_2,
    \dots, n_k)$. Write
    \[ \sigma = (a_{11} a_{12} \dots
    a_{1n_1})(a_{21} a_{22} \dots a_{2n_2}) \cdots
    (a_{k1} a_{k2} \dots a_{kn_k}) .\]
    Let $\tau \in S_n$. Then
    \begin{align*}
        \tau \sigma \tau^{-1}(\tau(a_{ij}))
        &= \tau \sigma(a_{ij}) \\
        &= \begin{cases}
            \tau(a_{ij}) & j < n_i \\
            \tau(a_{ii}) & j = n_i
        \end{cases}
    \end{align*}
    Thus 2 permutations of the same cycle type are
    conjugate.
\end{proof}

\noindent
For example,
\[ (14)(123)(14)^{-1} = (423) \]
\[ (1l)(1k)(1l) = (lk) .\]
Consider $S_4$: let $x \in S_4$. Recall $24 =
|S_4| = |\mathrm{ccl}_{S_4}(x)||C_{S_4}(x)|$
by Orbit-Stabiliser Theorem.

\begin{center}
    \begin{tabular}{c|c|c|c|c|c}
        example member $x$ & cycle type & size & sign & $|C_{S_4}(x)|$ & $C_{S_4}(x)$ \\
        \hline
        $e$ & $(1,1,1,1)$ & $1$ & $1$ & $24$ & $S_4$ \\
        $(12)(3)(4)$ & $(2,1,1)$ & $6$ & $-1$ & $4$ & $\langle (12), (34) \rangle \cong C_2 \times C_2$ \\
        $(123)(4)$ & $(3,1)$ & $8$ & $1$ & $3$ & $\langle (123) \rangle \cong C_3$ \\
        $(12)(34)$ & $(2,2)$ & $3$ & $1$ & $8$ & $\langle (1234), (12) \rangle \cong D_8$ \\
        $(1234)$ & $(4)$ & $6$ & $-1$ & $4$ & $\langle (1234) \rangle \cong C_4$
    \end{tabular}
\end{center}

\setcounter{customcorollary}{5}
\begin{corollary}
    The number of distinct conjugacy classes of
    $S_n$ is given by $p(n)$, the number of
    partitions of $n$ into positive integers,
    i.e. $n = n_1 + \cdots + n_k$ with $n_1 \ge
    n_2 \ge \cdots \ge n_k \ge 1$.
\end{corollary}

\noindent
However in $A_n$ conjugation is less clear.
Certainly
\[ \mathrm{ccl}_{A_n}(x) = \{gxg^{-1} : g \in
A_n\} \subseteq \{gxg^{-1} : g \in S_n\} =
\mathrm{ccl}_{S_n}(x) \]
since $A_n \le S_n$. \\
So if two elements are conjugate in $A_n$ they
have the same cycle type. But having the same
cycle type in $A_n$ does not guarantee being
conjugate. For example $(123)$ not conjugate to
$(132)$ in $A_4$. If $\tau (123) \tau^{-1} =
(132)$ then $\tau = (12)$, or $(32)$ or $(13)$,
none of which are in $A_4$. \myskip
Or consider $C_{A_4}((123)) = C_{S_4}((123)) \cap
A_4$. For example
\[ C_{S_4}((123)) = \langle (123) \rangle \le A_4 \]
So, $C_{A_4}((123)) = C_{S_4}((123))$
\[ \implies |\mathrm{ccl}((123))| =
\frac{|A_4|}{|C_{A_4}((123))|} = \frac{|S_4| /
2}{|C_{S_4}((123))|} =
\frac{|\mathrm{ccl}_{S_4}((123))|}{2} \]
So the conjugacy of 8 3-cycles in $S_4$ splits
into 2 conjugacy classes in $A_4$. \myskip
Key point: let $x \in A_n$. If $C_{A_n}(x) =
C_{5_n}(x)$
\[ \implies |\mathrm{ccl}_{A_n}(x)| =
\frac{|\mathrm{ccl}_{S_n}(x)|}{2} .\]
If $C_{A_n}(x) \le C_{S_n}(x)$, then $C_{S_n}(x)$
contains an odd permutation and
\[ |C_{A_n}(x)| = |C_{S_n}(x) \cap A_n| =
\frac{|C_{S_n}(x)|}{2} \]
(Sheet 2, Q4)
\[ \implies |\mathrm{ccl}_{A_n}(x)| =
|\mathrm{ccl}_{S_n}(x)| .\]

\begin{center}
    \begin{tabular}{c|c|c|c}
        example member $x$ & cycle type & $C_{A_4}(x)$ & size of conj class \\
        \hline
        $e$ & $(1,1,1,1)$ & $A_4$ & $1$ \\
        $(123)$ & $(3,1)$ & $\langle (123) \rangle$ & $4$ \\
        $(132)$ & $(3,1)$ & $\langle (132) \rangle$ & $4$ \\
        $(12)(34)$ & $(2,2)$ & $\{e, (12)(34), (13)(24), (14)(23)\} \cong C_2 \times C_2$ & $3$
    \end{tabular}
\end{center}

\begin{remark*}
    The number of elements in $S_n$ with $k_l$
    cycles of length $l$ is given by
    \[ \frac{n!}{\prod_l k_l! l^{k_l}} \]
    Think of cycles as trays, put in elements of
    $X = \{1, 2, \dots, n\}$. This gives $n!$
    options, but we've overcounted. Each cycle of
    length $l$ can be written $l$ ways, this gives
    $l^{k_l}$ factor. Also $k_l$ cycles of length
    $l$ can be permuted $k_l!$ ways.
\end{remark*}

\noindent
Let us consider $S_5$ (note $|S_5| = 120$).
\begin{center}
    \begin{tabular}{c|c|c|c|c|c}
        Example member $x$ & Cycle type & \# &
        $\mathrm{sgn}$ & $|C_{S_5}(x)|$ &
        $C_{S_5}(x)$ \\
        \hline
        $e$ & $(1,1,1,1,1)$ & 1 & 1 & 120 & $S_5$
        \\
        $(12)$ & $(2,1,1,1)$ & 10 & -1 & 12 &
        $\langle (12) \rangle \times
        \mathrm{Sym}\{3,4,5\} \cong C_2 \times
        S_3$
        \\
        $(12)(34)$ & $(2,2,1)$ & 15 & 1 & 8 &
        $\langle (1324),(12) \rangle \cong D_8$ \\
        $(123)$ & $(3,1,1)$ & 20 & 1 & 6 &
        $\langle (123),(45) \rangle \cong C_6$ \\
        $(123)(45)$ & $(32)$ & 20 & -1 & 6 &
        $\langle (123),(45) \rangle \cong C_6$ \\
        $(1234)$ & $(4,1)$ & 30 & -1 & 4 &
        $\langle (1234) \rangle \cong C_4$ \\
        $(12345)$ & $(5)$ & 24 & 1 & 5 & $\langle
        (12345) \rangle \cong C_5$
    \end{tabular}
\end{center}

\bigskip

\noindent
Now consider $A_5$ (note $|A_5| = 60$).
\begin{center}
    \begin{tabular}{c|c|c|c}
        Example member $x$ & Cycle type &
        $C_{A_5}(x)$ & $|\mathrm{ccl}_{A_5}(x)|$
        \\
        \hline
        $e$ & $(1,1,1,1,1)$ & $A_5$ & 1 \\
        $(12)(34)$ & $(2,2,1)$ & $\langle
        (12)(34),(13)(24) \rangle$ & 15 \\
        $(123)$ & $(3,1,1)$ & $\langle (123)
        \rangle$ & 20 \\
        $(12345)$ & $(5)$ & $\langle (12345)
        \rangle$ & 12 \\
        $(21345)$ & $(5)$ & $\langle (21345)
        \rangle$ & 12
    \end{tabular}
\end{center}
Recall a group is \emph{simple} if it has no
non-trivial proper normal subgroups, i.e. if only
normal subgroups are $\{e\}$ and $G$.

\setcounter{customtheorem}{9}
\begin{theorem}
    $A_5$ is a simple group.
\end{theorem}

\begin{proof}
    Suppose $N \normalsub A_5$. Then $N$ is a
    union of conjugacy classes (Sheet 3, question
    3(a)). Hence
    \[ |N| = 1 + 15a + 20b + 12c \]
    where $b, a \in \{0,1\}$ and $c \in
    \{0,1,2\}$. But by Lagrange's Theorem, $|N|
    \,\big|\, |A_5| = 60$. Only possibility is
    $|N| = 1$ or $|N| = 60$.
\end{proof}

\subsubsection*{Comments}

\begin{enumerate}[(i)]
    \item $A_5$ is the smallest non-abelian simple
        group.
    \item $A_n$ simple $\forall n \ge 5$ (GRM).
        But $A_4$ is not simple.
    \item Classification of finite simple
        groups exists, includes infinite families.
        \begin{itemize}
            \item $C_p$ for $p$ prime (only
                abelian simple groups).
            \item $A_n$ with $n \ge 5$.
            \item groups of `Lie type' (matrix
                groups)
            \item 26 sporadic groups (including
                the monster and baby monster)
        \end{itemize}
\end{enumerate}

\subsubsection*{Aside}

For example, number of cycles in $S_5$ of type
$(\bullet\,\,\bullet)(\bullet\,\,\bullet)$
so $k_2 = 2$, $k_1 = 1$.
\[ \# = \frac{q5!}{2!w^2 \cdot 1! \cdot 1} = 15 \]
For
$(\bullet\,\,\bullet\,\,\bullet)(\bullet\,\,\bullet)$
we have $k_3 = 1$, $k_2 = 1$
\[ \# = \frac{5!}{1!3^1 1!2^1} = 20 .\]

\newpage
\section{Matrix Groups}

Let $M_n(\RR)$ denote the set of all $n \times n$
matrices with entries in $\RR$. Define
\[ \mathrm{GL}_n(\RR) = \{A \in M_n(\RR) : \det A
\neq 0 \} \]

\begin{proposition}
    $\mathrm{GL}_n(\RR)$ is a group under matrix
    multiplication. It is called the \emph{general
    linear group}.
\end{proposition}

\begin{proof}
    Closure: $A, B \in \mathrm{GL}_n(\RR)$ clearly
    $AB \in M_n(\RR)$ and $\det(AB) = \det A \det
    B \neq 0$ so $AB \in \mathrm{GL}_n(\RR)$. \\
    Identity:
    \[ I_n =
    \begin{pmatrix}
        1 & 0 & \cdots & 0 & 0 \\
        0 & 1 & \cdots & 0 & 0 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & 1 & 0 \\
        0 & 0 & \cdots & 0 & 1
    \end{pmatrix}
    \in \mathrm{GL}_n(\RR) \]
    Inverse: $\det A \neq 0$ implies $A^{-1}$
    exists and $\det (A^{-1}) = \frac{1}{\det A}
    \neq 0$. \\
    Associative:
    \begin{align*}
        (A(BC))_{ij}
        &= A_{ix}(BC)_{xj} \\
        &= A_{ix}B_{xt}C_{tj} \\
        ((AB)C)_{ij}
        &= (AB)_{ix}C_{xj} \\
        &= A_{it}B_{tx}C_{xj}
    \end{align*}
\end{proof}

\begin{example*}
    We have that
    \[ \mathrm{GL}_2(\RR) = \left\{
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    : a, b, c, d \in \RR, ad - bc \neq 0 \right\} \]
    and we have
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    ^{-1} = \frac{1}{ad - bc}
    \begin{pmatrix}
        d & -b \\
        -c & a
    \end{pmatrix}
    \]
\end{example*}

\begin{proposition}
    \[ \det : \mathrm{GL}_n(\RR) \to (\RR
    \setminus \{0\}, \times) \qquad A \mapsto \det
    A \]
    is a surjective group homomorphism.
\end{proposition}

\begin{proof}
    Note $(\RR \setminus \{0\}, \times)$ is a
    group. Determinant is clearly a map to $(\RR
    \setminus \{0\}, \times)$. Need to
    check it's a group homomorphism
    \[ \det(AB) = \det A \cdot \det B \]
    And we need to show that it is surjective,
    which follows because given $r \in (\RR
    \setminus \{0\}, \times)$, let
    \[ A =
    \begin{pmatrix}
        r & 0 & \cdots & 0 & 0 \\
        0 & 1 & \cdots & 0 & 0 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & 1 & 0 \\
        0 & 0 & \cdots 0 & 1
    \end{pmatrix}
    \in \mathrm{GL}_n(\RR) \]
    and notice that $\det A = r$.
\end{proof}

By First Isomorphism Theorem
\[ \mathrm{Ker}(\det) \normalsub \mathrm{GL}_n(\RR) \]
and we can find that
\begin{align*}
    \mathrm{Ker}(\det)
    &= \{A \in \mathrm{GL}_n(\RR) : \det A = 1\}
    \\
    &= \mathrm{SL}_n(\RR)
\end{align*}
This is known as the \emph{special linear group}.
Furthermore, by First Isomorphism Theorem
\[ \mathrm{GL}_n(\RR) / \mathrm{SL}_n(\RR) \cong
(\RR \setminus \{0\}, \times) .\]

\begin{remark*}
    More generally we can define the general
    linear group and special linear group over any
    field. Examples of fields: $\RR$, $\CC$,
    $\QQ$, $\FF_p$ where
    \[ \FF_p = (\{0,1,2, \dots, p - 1\}, +_p,
    \times_p) \]
    for some prime $p$. Note that
    $\mathrm{GL}_n(\FF_p)$ and
    $\mathrm{SL}_n(\FF_p)$ are finite groups.
\end{remark*}

\noindent
What is $|\mathrm{GL}_3(\FF_p)|$? Non-zero
determinant means we need linearly independent
columns. So the number of choices for first column
is $p^3 - 1$ (any choice is fine except
$(0, 0, 0)$). Second column is not a multiple of
first, so number of choices for second column is
$p^3 - p$. (Note that the zero vector is a
multiple of the first column). Third column not in
space spanned by first two columns, this space has
size $p^2$ (consider $\alpha c_1 + \beta c_2$,
$\alpha, \beta \in \FF_p$). So number of choices
for third column is $p^3 - p^2$. So
\[ |\mathrm{GL}_3(\FF_p)| = (p^3 - 1)(p^2 - p)(p^3
- p^2) \]
We can still consider
\[ \det : \mathrm{GL}_3(\FF_p) \to (\FF_p
\setminus \{0\}, \times) \qquad A \mapsto \det A \]
Note $(\FF_p \setminus \{0\}, \times)$ is a group.

\begin{proof}
    Closure, identity and associativity can
    all easily be verified. Let $a \in \FF_p \setminus
    \{0\}$, by Bezout's Theorem, there exists $x, y$
    such that $ax + py = 1$. Then we have $ax \equiv 1
    \pmod{p}$. Choose $\ol{x} \equiv x \pmod{p}$ with
    $1 \le \ol{x} \le p - 1$. So $a^{-1} \equiv
    x$.
\end{proof}

Determinant is a surjective homomorphism
to $(\FF_p \setminus \{0\}, \times)$ so by
First Isomorphism Theorem:
\[ |\mathrm{GL}_3(\FF_p)| /
|\mathrm{SL}_2(\FF_p)| = p - 1 \]
\[ \implies |\mathrm{SL}_3(\FF_p)| =
\frac{(p^3 - 1)(p^2 - p)(p^3 - p^2)}{p - 1} \]

\subsubsection*{Actions of $\mathrm{GL}_n(\CC)$}

\begin{enumerate}[(i)]
    \item Let $\CC^n$ denote vectors of length $n$
        with entries in $\CC$:
        \[ \mathrm{GL}_n(\CC) \times \CC^n \to
        \CC^n \qquad (A, \bf{v}) \mapsto A\bf{v} \]
        Note $I\bf{v} = \bf{v}$, $(AB)\bf{v} =
        A(B(\bf{v}))$. This action is faithful:
        \[ A\bf{v} = \bf{v} \,\,\forall \bf{v}
        \in \CC^n \implies A = I_n \]
        (consider multiplying $A$ by
        $(1,0,\dots,0)$, $(0,1,\dots,0)$ etc) The
        action has two orbits:
        \[ \mathrm{Orb}_{\mathrm{GL}_n(\CC)}
        (\bf{0}) = \{\bf{0}\} \qquad \bf{0} =
        \begin{pmatrix}
            0 \\
            0 \\
            \vdots \\
            0
        \end{pmatrix}
        \]
        and for $\bf{v} \neq 0$ we have:
        \[
        \mathrm{Orb}_{\mathrm{GL}_n(\CC)}(\bf{v})
        = \CC^n \setminus \{\bf{0}\} \]
        because given $\bf{w} \neq \bf{0}$ there exists
        $A \in \mathrm{GL}_n(\CC)$ such that
        $A\bf{v} = \bf{w}$.
    \item Conjugation action of
        $\mathrm{GL}_n(\CC)$ on $M_n(\CC)$
        \[ \mathrm{GL}_n(\CC) \times M_n(\CC) \to
        M_n(\CC) \qquad (P, A) \mapsto PAP^{-1} \]
        Note:
        \begin{align*}
            PQ(A)
            &= PQA(PQ)^{-1} \\
            &= PQAQ^{-1}P^{-1} \\
            &= P(Q(A))
        \end{align*}
\end{enumerate}
        \begin{remark*}
            Matrices $A$ and $B$ are conjugate if
            they represent the same linear map. If
            $PAP^{-1} = B$, then $P$ represents a
            change of basis matrix (see linear
            algebra next year). For example
            \[ e_1 =
            \begin{pmatrix}
                1 \\
                0
            \end{pmatrix}
            \quad e_2 =
            \begin{pmatrix}
                0 \\
                1
            \end{pmatrix}
            \]
            \[ A : e_1 \mapsto 2e_1 \qquad e_2
            \mapsto 3e_2 \]
            \[ A =
            \begin{pmatrix}
                2 & 0 \\
                0 & 3
            \end{pmatrix}
            \]
            Let
            \[ P : e_1 \mapsto e_2, \quad
            e_2 \mapsto e_1 \]
            change of basis
            \[ P =
            \begin{pmatrix}
                0 & 1 \\
                1 & 0
            \end{pmatrix}
            = p^{-1} \]
            Then
            \begin{align*}
                PAP^{-1}
                &=
                \begin{pmatrix}
                    0 & 1 \\
                    1 & 0
                \end{pmatrix}
                \begin{pmatrix}
                    2 & 0 \\
                    0 & 3
                \end{pmatrix}
                \begin{pmatrix}
                    0 & 1 \\
                    1 & 0
                \end{pmatrix}
                \\
                &=
                \begin{pmatrix}
                    0 & 3 \\
                    2 & 0
                \end{pmatrix}
                \begin{pmatrix}
                    0 & 1 \\
                    1 & 0
                \end{pmatrix}
                \\
                &=
                \begin{pmatrix}
                    3 & 0 \\
                    0 & 2
                \end{pmatrix}
            \end{align*}
            i.e. $e_2 \mapsto 3e_2$ and $e_1
            \mapsto 2e_1$. We will use the
            following result from Vectors and
            Matrices when investigating M\"obius
            groups.
        \end{remark*}

\begin{result*}
    Let $A \in M_2(\CC)$ and consider conjugation
    action of $\mathrm{GL}_2(\CC)$ on $M_2(\CC)$.
    Then precisely one of the following occurs:
    \begin{enumerate}[(i)]
        \item the orbit of $A$ contains a diagonal
            matrix
            \[
            \begin{pmatrix}
                \lambda & 0 \\
                0 & \mu
            \end{pmatrix}
            \]
            with $\lambda \neq \mu$.
        \item the orbit of $A$ is
            \[
            \begin{pmatrix}
                \lambda & 0 \\
                0 & \lambda
            \end{pmatrix}
            = \lambda I \]
            for some $\lambda$.
        \item the orbit of $A$ contains a matrix
            \[
            \begin{pmatrix}
                \lambda & 1 \\
                0 & \lambda
            \end{pmatrix}
            \]
            for some $\lambda$.
    \end{enumerate}
\end{result*}

\begin{proof}
    See Vectors and Matrices but essentially
    \begin{enumerate}[(i)]
        \item In this case $A$ has 2 distinct
            eigenvalues $\lambda \neq \mu$, take a
            basis consisting of an eigenvector for
            $\lambda$ and an eigenvector for
            $\mu$. Distinct pairs give distinct
            orbits.
        \item $A = \lambda I$, eigenvalues
            $\lambda$, $\lambda$, 2 linearly
            independent eigenvectors.
        \item In this case $A$ has a repeated
            eigenvalue, but just one linearly
            independent eigenvector.
    \end{enumerate}
\end{proof}

\bigskip

\noindent
Recall if $A \in M_(\RR)$, $A^\top$ is defined by
$(A^\top)_{ij} = A_{ji}$, i.e. the $ij$-th entry
of $A^\top$ is $ji$-th entry of $A$:
\[ A =
\begin{pmatrix}
    2 & 4 \\
    3 & 5
\end{pmatrix}
\qquad A^\top =
\begin{pmatrix}
    2 & 3 \\
    4 & 5
\end{pmatrix}
\]
\begin{note*}
    \begin{enumerate}[(i)]
        \item We have $(AB)^\top = B^\top A^\top$
            because
            \[ [(AB)^\top]_{ij} = (AB)_{ji} =
            A{jk}B_{ki} \]
            \[ [B^\top A^\top]_{ij} =
            B^\top_{ik}A^\top_{kj} = B_{ki} A_{jk} \]
        \item $AA^\top = I \iff A^\top A = I$ and
            hence
            \[ A^\top A = A^{-1}A A^\top A =
            A^{-1}A = I \]
        \item $(A^\top)^{-1} = (A^{-1})^\top$
            since
            \begin{align*}
                I_n
                &= (AA^{-1})^\top \\
                &= (A^{-1})^\top A^\top
            \end{align*}
        \item $\det(A^\top) = \det A$.
    \end{enumerate}
\end{note*}

\[ \mathrm{O}_n(\RR) = \{A \in M_n(\RR) : A^\top A = I\} \]
(So columns of $A$ form an orthonormal basis for $\RR^n$).

\setcounter{customproposition}{9}
\begin{proposition}
    $\mathrm{O}_n(\RR)$ is a subgroup of
    $\mathrm{GL}_n(\RR)$ called the
    \emph{orthogonal group}.
\end{proposition}

\begin{proof}
    \begin{align*}
        1 
        &= \det(A^\top A) \\
        &= \det(A^\top) \det(A) \\
        &= (\det A)^2 \\
        \implies \det A \\
        &\neq 0
    \end{align*}
    Hence $\mathrm{O}_n(\RR)$ is a subset of
    $\mathrm{GL}_n(\RR)$; associativity is
    inherited.
    \begin{itemize}
        \item $I_n =
            \begin{pmatrix}
                1 & \cdots & 0 \\
                \vdots & \ddots & \vdots \\
                0 & \cdots & 1
            \end{pmatrix}
            \in \mathrm{O}_n(\RR)$
        \item closure: $A, B \in \mathrm{O}_n(\RR)$,
            \begin{align*}
                (AB)^\top (AB)
                &= B^\top A^\top A B \\
                &= B^\top B \\
                &= I \\
                \implies B
                &\in \mathrm{O}_n(\RR)
            \end{align*}
        \item inverse: $A^\top A = I_n \implies
            A^\top = A^{-1}$ and $A^\top \in
            \mathrm{O}_n(\RR)$ since $(A^\top)^\top = A$
            and $AA^\top = I$.
    \end{itemize}
\end{proof}

\noindent
Note $1 = (\det A)^2 \implies \det A = \pm 1$ if
$A \in \mathrm{O}_n(\RR)$. So, $\mathrm{Det} :
\mathrm{O}_n(\RR) \to
(\{\pm 1\}, \times)$, $A \mapsto \det A$ is a
surjective homomorphism, as
\[
\begin{pmatrix}
    -1 & 0 & \cdots & 0 \\
    0 & 1 & \cdots & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & 1
\end{pmatrix}
\in \mathrm{O}_n(\RR) \]
So
\[ \mathrm{Ker}(\mathrm{Det}) = \{A \in \mathrm{O}_n(\RR) :
\det A = 1\} = \mathrm{SO}_n(\RR) \normalsub
\mathrm{O}_n(\RR) \]
By First Isomorphism Theorem:
\[ \mathrm{O}_n(\RR) / \mathrm{SO}_n(\RR) \cong
C_2 \]

\setcounter{customlemma}{19}
\begin{lemma}
    Let $A \in \mathrm{O}_n(\RR)$ and $\bf{x},
    \bf{y} \in \RR^n$. Then
    \begin{enumerate}[(i)]
        \item $A \bf{x} \cdot A \bf{y} = \bf{x}
            \cdot \bf{y}$
        \item $|A \bf{x}| = |\bf{x}|$
    \end{enumerate}
    So $A$ is an isometry (distance preserving
    map) of Euclidean space $\RR^n$.
\end{lemma}

\begin{proof}
    \begin{enumerate}[(i)]
        \eqitem
            \begin{align*}
                A \bf{x} \cdot A \bf{y}
                &= (A \bf{x})^\top (A \bf{y}) \\
                &= \bf{x}^\top A^\top A \bf{y} \\
                &= \bf{x}^\top \bf{y} \\
                &= \bf{x} \cdot \bf{y}
            \end{align*}
        \item \[ |A \bf{x}|^2 = A \bf{x} \cdot A
            \bf{x} = \bf{x} \cdot \bf{x} =
            |\bf{x}|^2 \]
    \end{enumerate}
\end{proof}

\noindent
Note by (ii) if $\lambda$ an eigenvalue of $A$,
then $A \bf{x} = \lambda \bf{x}$
\[ \implies |\lambda \bf{x}| = |\bf{x}| \]
i.e. $|\lambda| = 1$.

\subsubsection*{In 2 dimensions}

Let
\[ A =
\begin{pmatrix}
    a & b \\
    c & d
\end{pmatrix}
\in \mathrm{GL}_2(\RR) \]
\[ I = AA^\top =
\begin{pmatrix}
    a & b \\
    c & d
\end{pmatrix}
\begin{pmatrix}
    a & c \\
    b & d
\end{pmatrix}
\]
\[ \implies 1 = a^2 + b^2 = c^2 + d^2 \]
\[ 0 = ac + bd .\]
\[ I = A^\top A =
\begin{pmatrix}
    a & c \\
    b & d
\end{pmatrix}
\begin{pmatrix}
    a & b \\
    c & d
\end{pmatrix}
\]
\[ \implies 1 = a^2 + c^2 = b^2 + d^2 \]
\[ 0 = ab + cd \]
For $0 \le \theta < 2\pi$ let
\[
\begin{pmatrix}
    a \\
    c
\end{pmatrix}
=
\begin{pmatrix}
    \cos\theta \\
    \sin\theta
\end{pmatrix}
\qquad \text{so} \qquad
\begin{pmatrix}
    b \\
    d
\end{pmatrix}
= \pm
\begin{pmatrix}
    -\sin\theta \\
    \cos\theta
\end{pmatrix}
\]
\ul{First case}:
\[ A =
\begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\]
$\det A = 1$
\[ A
\begin{pmatrix}
    x \\
    y
\end{pmatrix}
=
\begin{pmatrix}
    \cos\theta x & -\sin\theta y \\
    \sin\theta x & \cos\theta y
\end{pmatrix}
\]
$A$ represents a rotation. \\
Let $z = x + iy$ then
\[ e^{i\theta} z = (\cos\theta x - \sin\theta y) +
i(\sin\theta x + \cos\theta y) \]
All elements of $\mathrm{SO}_2(\RR)$ are of this
form. \myskip
\ul{Second case}
\[ A =
\begin{pmatrix}
    \cos\theta & \sin\theta \\
    \sin\theta & -\cos\theta
\end{pmatrix}
\]
$\det A = -1$
\[ A
\begin{pmatrix}
    x \\
    y
\end{pmatrix}
=
\begin{pmatrix}
    \cos\theta x & \sin\theta y \\
    \sin\theta x & -\cos\theta y
\end{pmatrix}
\]
\[ e^{i\theta} \ol{z} = (\cos\theta x + \sin\theta
y) + i(\sin\theta x - \cos\theta y) \]
What are the fixed points?
\begin{align*}
    z = e^{i\theta} \ol{z}
    &\iff e^{-\theta / 2} z = e^{i\theta/2} \ol{z}
    \\
    &\iff e^{-i\theta/2} z = t \in \RR \\
    &\iff z = e^{i\theta/2} t
\end{align*}
hence a reflection in line $te^{i\theta/2}$. \\
All elements of $\mathrm{O}_2(\RR) \setminus
\mathrm{SO}_2(\RR)$ are of this form. \myskip
So,
\[ \mathrm{O}_2(\RR) = \mathrm{SO}_2(\RR) \cup
\begin{pmatrix}
    1 & 0 \\
    0 & -1
\end{pmatrix}
\mathrm{SO}_2(\RR) \]
Note any element of $\mathrm{O}_2(\RR)$ is a
product of at most two reflections. Since if $A
\in \mathrm{SO}_2(\RR)$ then
\[ A =
\left( A
\begin{pmatrix}
    1 & 0 \\
    0 & -1
\end{pmatrix}
\right)
\begin{pmatrix}
    1 & 0 \\
    0 & -1
\end{pmatrix}
\]

\subsubsection*{3 dimensions}

\setcounter{customproposition}{10}
\begin{proposition}
    Let $A \in \mathrm{SO}_3(\RR)$. Then $A$ has
    an eigenvector with eigenvalue 1.
\end{proposition}

\begin{proof}
    \begin{align*}
        \det(A - I)
        &= \det(A - AA^\top) \\
        &= \det A \det(I - A^\top) \\
        &= \det((I - A)^\top) \\
        &= \det(I - A) \\
        &= (-1)^3 \det(A - I) \\
        &= -\det(A - I)
    \end{align*}
    hence $\det(A - I) = 0$ and $A$ has eigenvalue
    1.
\end{proof}

\noindent
Alternatively consider $\chi_A(x)$ the
characteristic polynomial of $A$, it is a cubic in
$\RR$. Thus has a real root, $\lambda = 1$ or
$\lambda = -1$. But the other eigenvalues are
either a complex conjugate pair, then $\lambda =
1$ or all are real either $1, -1, -1$ or $1, 1,
1$.

\setcounter{customtheorem}{10}
\begin{theorem}
    Let $A \in \mathrm{SO}_3(\RR)$ then $A$ Is
    conjugate to a matrix of the form
    \[
    \begin{pmatrix}
        \cos\theta & -\sin\theta & 0 \\
        \sin\theta & \cos\theta & 0 \\
        0 & 0 & 1
    \end{pmatrix}
    \]
    for some $\theta \in [0, 2\pi]$. In
    particular, $A$ is a rotation round an axis
    through the origin.
\end{theorem}

\begin{proof}
    By proposition 11, there is a $\bf{v} \in
    \RR^3$ with $A \bf{v} = \bf{v}$, and we can
    assume $|\bf{v}| = 1$. Let $\{e_1, e_2, e_3\}$
    be the standard orthonormal basis for $\RR^3$.
    There exists $P \in \mathrm{SO}_3(\RR)$ such
    that $P \bf{v} = e_3$. So $PAP^{-1}(e_3) =
    e_3$ and for $\pi$ plane perpendicular to
    $e_3$ then $PAP^{-1}(\pi)$ perpendicular to
    $e_3$. So,
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/4a7dbcf47e0b11ec.png}
    \end{center}
    $\det PAP^{-1} = \det A = 1$, so $\det Q = 1$,
    $Q^\top Q = I$. So
    \[ Q =
    \begin{pmatrix}
        \cos\theta & -\sin\theta \\
        \sin\theta & \cos\theta
    \end{pmatrix}
    \]
    for some $\theta$ as required.
\end{proof}

\noindent
Suppose $\bf{r}$ is a reflection in a plane $\pi$
through $0$. Let $\bf{n}$ be unit vector
perpendicular to $\pi$. Then
\[ r(\bf{x}) = \bf{x} - 2(\bf{x} \cdot \bf{n}) \bf{n} \]
\[ \bf{n} \mapsto -\bf{n} \]
$\pi$ fixed. So $\bf{r}$ is conjugate to
\[
\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
\end{pmatrix}
\in \mathrm{O}_3(\RR) \]
\[ \mathrm{O}_3(\RR) = \mathrm{SO}_3(\RR) \cup
\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
\end{pmatrix}
\mathrm{SO}_3(\RR) \]

\setcounter{customtheorem}{12}
\begin{theorem}
    Any element of $\mathrm{O}_3(\RR)$ is a
    product of at most 3 reflections.
\end{theorem}

\begin{proof}
    Let $\{e_1, e_2, e_3\}$ be standard
    orthonormal basis for $\RR^3$. Let $A \in
    \mathrm{O}_3(\RR)$. Then
    \[ |Ae_3| = |e_3| = 1 ,\]
    since $A$ is an isometry. So there exists a
    reflection $r_1$ such that
    \[ r_1A(e_3) = e_3 .\]
    Let $\pi = \langle e_1, e_2 \rangle$ (the
    plane perpendicular to $e_3$). Then $r_1A(\pi)
    = \pi$. There exists a reflection $r_2$ such
    that
    \[ r_2(e_3) = e_3, \qquad r_2(r_1A(e_2)) = e_2
    .\]
    So $r_2r_1A$ fixes $e_2$ and $e_3$. So
    $r_2r_1A(e_1) = \pm e_1$. If $e_1 = e_1$, set
    $r_3 = \id$. If $e_1 = -e_1$, let $r_3$ be
    reflection in plane perpendicular to $e_1$. So
    $r_3r_2r_1A$ fixes $e_1, e_2, e_3$, so
    \[ r_3r_2r_1 A = \id \]
    \[ \implies A = r_1^{-1}r_2^{-1}r_3^{-1} =
    r_1r_2r_3 .\]
\end{proof}

\myskip
Alternatively, any element in $\mathrm{S0}_2(\RR)$
is a product of at most 2 reflections, via
2-dimensional case. Thus any element of
\[
\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
\end{pmatrix}
\mathrm{SO}_3(\RR) \]
is a product of at most 3 reflections. Note we do
need 3, for example consider
\[
\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & -1
\end{pmatrix}
\]

\newpage
\section{M\"obius Groups}

A M\"obius transformation (or map) is a function
of a complex variable $z$ that can be written in
the form
\[ f(z) = \frac{az + b}{cz + d} \]
for some $a, b, c, d \in \CC$ with $ad - bc \neq
0$. Why $ad - bc \neq 0$?
\[ f(z) - f(w) = \frac{(ad - bc)(z - w)}{(cz +
d)(cw + d)} .\]
So, $ad - bc = 0$ implies $f$ constant (not
interesting), and $ad - bc \neq 0$ implies $f$
injective. When does $f(z) = g(z)$? \\
Suppose there exists at least 3 values of $z$ in
$\CC$ such that
\[ \frac{az + b}{cz + d} = \frac{\alpha z +
\beta}{\gamma z + \delta} \]
$ad - bc \neq 0$, $\alpha \delta - \beta \gamma
\neq 0$. Then there exists $\lambda \neq 0$,
$\lambda \in \CC$ such that
\[
\begin{pmatrix}
    \alpha & \beta \\
    \gamma & \delta
\end{pmatrix}
= \lambda
\begin{pmatrix}
    a & b \\
    c & d
\end{pmatrix}
\]
Since, we have 3 distinct values of $z$ for which
\[ (az + b)(\gamma z + \delta) = (\alpha z +
\beta)(cz + d) \]
so these quadratics are identical. Hence
\[ a \gamma = \alpha c, \qquad b \delta = \beta d \]
\[ a \delta + b \gamma = \alpha d + \beta c \]
Let $\mu = a \delta - \beta c = \alpha d - b
\gamma$ (so $\mu^2 = (ad - bc)(\alpha \delta -
\beta \gamma) \neq 0$). Then
\[
\begin{pmatrix}
    d & -b \\
    -c & a
\end{pmatrix}
\begin{pmatrix}
    \alpha & \beta \\
    \gamma & \delta
\end{pmatrix}
=
\begin{pmatrix}
    \mu & 0 \\
    0 & \mu
\end{pmatrix}
\]
\[ \implies
\begin{pmatrix}
    \alpha & \beta \\
    \gamma & \delta
\end{pmatrix}
= \frac{\mu}{ad - bc}
\begin{pmatrix}
    a & b \\
    c & d
\end{pmatrix}
\]

\myskip
Problem: $f$ is not defined at $z = -\frac{d}{c}$.
We would like $f \left( - \frac{d}{c} \right) =
\infty$. We consider $f$ defined on $\CC \cup
\{\infty\} = \CC_\infty$, the extended complex
plane. So if
\[ f(z) = \frac{az + b}{cz + d} ,\]
domain is now $\CC_\infty$; $c \neq 0$; $f(\infty)
= \frac{a}{c}$, $f \left( - \frac{d}{c} \right) =
\infty$. For $c = 0$; $f(\infty) = \infty$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/5e0b8c2c7e3f11ec.png}
\end{center}
(Riemann Sphere and stereographic projection.)

\begin{theorem}
    The set $\mathcal{M}$ of all M\"obius maps on
    $\CC_\infty$ is a group under composition. It
    is a subgroup of $\mathrm{Sym}(\CC_\infty)$.
\end{theorem}

\begin{proof}
    \begin{itemize}
        \item composition of maps is associative
        \item $I(z) = z \in \mathcal{M}$.
        \item closure: Let
            \[ f(z) = \frac{az + b}{cz + d},
            \qquad g(z) = \frac{\alpha z +
            \beta}{\gamma z + \delta} \]
            Suppose $c \neq 0$, $\delta \neq 0$.
            First suppose $z \in \CC \setminus \{
            -\delta / \gamma\}$. Then
            \begin{align*}
                f(g(z))
                &= \frac{a \left( \frac{\alpha z +
                \beta}{\gamma z + \delta} \right)
                + b}{c \left( \frac{\alpha z +
                \beta}{\gamma z + \delta} \right)
                + d} \\
                &= \frac{(a\alpha + b\gamma)z +
                (a\beta + b\delta)}{(c\alpha +
                d\gamma) + (c\beta + \delta d)}
                \in \mathcal{M}
            \end{align*}
            since
            \[ (a\alpha + b\gamma)(c\beta + \delta
            d) - (a\beta + b\delta)(c\alpha +
            d\gamma) = (ad - bc)(\alpha \delta -
            \beta \gamma) \neq 0 .\]
            Also, $f \left( g \left( -
            \frac{\delta}{\gamma} \right)  \right)
            = f(\infty) = \frac{a}{c}$. And
            \begin{align*}
                \frac{(a\alpha + b\gamma) \left(
                -\frac{\delta}{\gamma} \right) +
                (a\beta + b\delta)}{(c\alpha +
                d\gamma) \left(
                -\frac{\delta}{\gamma} \right) +
                (c\beta + \delta d)}
                &= \frac{a\alpha \left( -
                \frac{\delta}{\gamma} \right) +
                \alpha \beta}{c \alpha \left(
                -\frac{\delta}{\gamma} \right) +
                c\beta} \\
                &= \frac{a}{c}
            \end{align*}
            Need to check $c = 0$ separately.
        \item inverses: For some $a, b, c, d$ with
            $ad - bc \neq 0$, let
            \[ f(z) = \frac{az + b}{cz + d} \qquad
            \text{and} \qquad f^*(z) = \frac{dz -
            b}{-cz + a} \]
            Then $f(f^*(z)) = z = f^*(f(z))$ for
            $z \neq -\frac{d}{c}, -\frac{a}{c},
            \infty$. These are cases are ok. If $c
            = 0$ then
            \[ f(f^*(\infty)) = f(\infty = \infty
            = f^*(f(\infty)) .\]
    \end{itemize}
\end{proof}

\begin{theorem}
    \[ \frac{\mathrm{GL}_2(\CC)}{Z} \cong
    \mathcal{M} \]
    where
    \[ Z = \{
    \begin{pmatrix}
        \lambda & 0 \\
        0 & \lambda
    \end{pmatrix}
    : \lambda \in \CC \setminus \{0\}\} .\]
\end{theorem}

\begin{proof}
    We construct a surjective homomorphism from
    $\mathrm{GL}_2(\CC)$ onto $\mathcal{M}$ with
    kernel $Z$. \\
    Let $\phi : \mathrm{GL}_2(\CC) \to \mathcal{M}$
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    \mapsto f(z) = \frac{az + b}{cz + d} .\]
    Note $\phi$ a homomorphism:
    \[ f(z) = \frac{az + b}{cz + d}, \qquad g(z) =
    \frac{\alpha z + \beta}{\gamma z + \delta} .\]
    \begin{align*}
        \phi \left(
        \begin{pmatrix}
            a & b \\
            c & d
        \end{pmatrix}
        \right) \phi \left(
        \begin{pmatrix}
            \alpha & \beta \\
            \gamma & \delta
        \end{pmatrix}
        \right) (z)
        &= f \circ g(z) \\
        &= \frac{(a\alpha + b\gamma)z + (a\beta +
        b\delta}{(c\alpha + d\gamma)z + (c\beta +
        \delta d)} \\
        &= \phi \left(
        \begin{pmatrix}
            a & b \\
            c & d
        \end{pmatrix}
        \begin{pmatrix}
            \alpha & \beta \\
            \gamma & \delta
        \end{pmatrix}
        \right)
    \end{align*}
    Clearly $\phi$ surjective.
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    \in \mathrm{Ker} \,\,\phi \]
    if and only if $\frac{az + b}{cz + d} = z
    \,\,\forall\,\, z \in \CC_\infty$. Note
    \[ z = \infty \implies c = 0 \]
    \[ z = 0 \implies b = 0 \]
    \[ z = 1 \implies a = d \]
    \[ \implies \mathrm{Ker}\,\,\phi = z \]
    Finally apply First Isomorphism Theorem.
\end{proof}

\setcounter{customcorollary}{6}
\begin{corollary}
    \[ \frac{\mathrm{SL}_2(\CC)}{\{\pm I\}} \cong
    \mathcal{M} .\]
\end{corollary}

\begin{proof}
    Restrict $\phi$ to $\mathrm{SL}_2(\CC)$
    \[ \phi : \mathrm{SL}_2(\CC) \to \mathcal{M} \]
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    \mapsto \frac{az + b}{cz + d} .\]
    We require $\phi$ to be surjective:
    \[ f(z) = \frac{az + b}{cz + d} =
    \frac{\left( \frac{a}{(ad - bc)^{1/2}} \right)
    z + \frac{b}{(ad - bc)^{1/2}}}{\left(
    \frac{c}{(ad - bc)^{1/2}} \right) z +
    \frac{d}{(ad - bc)^{1/2}}} .\]
    And $\mathrm{Ker}\,\,\phi = \{\pm I\}$.
\end{proof}

\setcounter{customproposition}{12}
\begin{proposition}
    Every M\"obius map can be written as a
    somposition of maps of the following forms:
    \begin{enumerate}[(i)]
        \item $z \mapsto az$, $a \neq 0$;
            represents a dilation or rotation
        \item $z \mapsto z + b$; a translation
        \item $z \mapsto \frac{1}{z}$; inversion.
    \end{enumerate}
\end{proposition}

\begin{proof}
    Let $f(z) = \frac{az + b}{cz + d}$. If $c =
    0$;
    \[ z \mapsto \left( \frac{a}{d} \right) z
    \to \mapsto \left( \frac{a}{d} \right) z +
    \left( \frac{b}{d} \right) \]
    $f_1$ is type (i), $f_2$ is type (ii). We can
    write $f = f_2 \circ f_1$. If $c \neq 0$,
    write
    \begin{align*}
        f(z)
        &= \frac{az + b}{cz + d} \\
        &= \frac{\left( \frac{a}{c} \right) z +
        \left( \frac{b}{c} \right)}{z + \left(
        \frac{d}{c} \right)} \\
        &= \frac{a}{c} + \frac{\left( \frac{-ad +
        bc}{c^2} \right)}{\left( z + \frac{d}{c}
        \right)} \\
        &= A + \frac{B}{z + \frac{d}{c}}
    \end{align*}
    \[ z \stackrel{\text{(ii)}}{\mapsto} z +
    \frac{d}{c} \stackrel{\text{(iii)}}{\mapsto}
    \frac{1}{z + \frac{d}{c}}
    \stackrel{\text{(i)}}{\mapsto} \frac{B}{z +
    \frac{d}{c}} \stackrel{\text{(ii)}}{\mapsto} A
    + \frac{B}{z + \frac{d}{c}} .\]
    Now we can write $f = f_4 \circ f_3 \circ f_2
    \circ f_1$.
\end{proof}

\setcounter{customdefinition}{21}
\begin{definition}
    A group $G$ acts \emph{triply transitively} on
    a set $X$ if given $x_1, x_2, x_3 \in X$ all
    distinct and $y_1, y_2, y_3 \in X$ all
    distinct, there exists $g \in G$ such that
    $g(x_i) = y_i$, for $i = 1, 2, 3$. \\
    A group $G$ acts \emph{sharply triply
    transitively} if such a $g$ is unique.
\end{definition}

\begin{theorem}
    The action of $\mathcal{M}$ on $\CC_\infty$ is
    sharply triply transitive.
\end{theorem}

\begin{proof}
    Label first triple $\{z_0, z_1, z_\infty\}$
    and second triple $\{\omega_0, \omega_1,
    \omega_\infty\}$. We construct $g \in
    \mathcal{M}$ such that
    \begin{align*}
        g : z_0
        &\mapsto 0 \\
        z_1
        &\mapsto 1 \\
        z_\infty
        &\mapsto \infty
    \end{align*}
    First suppose $z_0, z_1, z_\infty \neq
    \infty$
    \[ g(z) = \frac{(z - z_0)(z_1 - z_\infty)}{(z
    - z_\infty)(z_1 - z_0)} \]
    check: ``$ad - bc$'' $= (z_0 - z_\infty)(z_1 -
    z_\infty)(z_1 - z_0) \neq 0$. If $z_\infty =
    \infty$:
    \[ g(z) = \frac{(z - z_0)}{(z_1 - z_0)} \]
    If $z_1 = \infty$:
    \[ g(z) = \frac{(z - z_0)}{(z - z_\infty)} \]
    If $z_0 = \infty$:
    \[ g(z) = \frac{(z_1 - z_\infty)}{(z -
    z_\infty)} .\]
    Similarly find $h$ such that
    \begin{align*}
        h : \omega_0
        &\mapsto 0 \\
        \omega_1
        &\mapsto 1 \\
        \omega_\infty
        &\mapsto \infty
    \end{align*}
    Then $f = h^{-1}g : z_i \mapsto \omega_i$ as
    required. Now to prove uniqueness. Suppose
    $f' : z_i \mapsto \omega_i$. Then $f^{-1}f' :
    z_i \mapsto z_i$. Let $g$ be as above, then
    \begin{align*}
        gf^{-1}f' g^{-1} : 0
        &\mapsto 0 \implies b = 0 \\
        1
        &\mapsto 1 \implies a = d \\
        \infty
        &\mapsto \infty \implies c = 0
    \end{align*}
    \[ \implies gf^{-1}f'g^{-1} = \id \]
    \[ \implies f^{-1}f' = \id \]
    \[ \implies f = f' .\]
\end{proof}

\myskip
So, the image of just three points determines the
map.

\subsubsection*{Conjugacy classes in $\mathcal{M}$}

Recall $\phi : \mathrm{GL}_2(\CC) \surjto
\mathcal{M}$. Suppose $A$, $B$ conjugate in
$\mathrm{GL}_2(\CC)$, i.e. there exists $P \in
\mathrm{GL}_2(\CC)$ such that
\[ PAP^{-1} = B \]
then
\begin{align*}
    \phi(P)\phi(A)\phi(P)^{-1}
    &= \phi(PAP^{-1}) \\
    &= \phi(B) \in \mathcal{B}
\end{align*}
i.e. $\phi(A)$ and $\phi(B)$ are conjugate in
$\mathcal{M}$. Use knowledge of conjugacy classes
in $\mathrm{GL}_2(\CC)$.
\begin{enumerate}[(i)]
    \item For some $\lambda \neq \mu$, $\lambda
        \neq 0 \neq \mu$
        \[
        \begin{pmatrix}
            \lambda & 0 \\
            0 & \mu
        \end{pmatrix}
        \]
        \[ \phi \left(
        \begin{pmatrix}
            \lambda & 0 \\
            0 & \mu
        \end{pmatrix}
        \right) = f \]
        $f(z) = \nu z$, $\nu \neq 0, 1$.
    \item For some $\lambda \neq 0$,
        \[
        \begin{pmatrix}
            \lambda & 0 \\
            0 & \lambda
        \end{pmatrix}
        \]
        \[ \phi \left(
        \begin{pmatrix}
            \lambda & 0 \\
            0 & \lambda
        \end{pmatrix}
        \right) = \id .\]
    \item For some $\lambda \neq 0$,
        \[
        \begin{pmatrix}
            \lambda & 1 \\
            0 & \lambda
        \end{pmatrix}
        \]
        \[ \phi \left(
        \begin{pmatrix}
            \lambda & 1 \\
            0 & \lambda
        \end{pmatrix}
        \right) = f \]
        $f(z) = \frac{\lambda z + 1}{\lambda} = z
        + \frac{1}{\lambda}$, i.e.
        \[ f = \phi \left(
        \begin{pmatrix}
            1 & \frac{1}{\lambda} \\
            0 & 1
        \end{pmatrix}
        \right) \]
        And it's conjugate to
        \[
        \begin{pmatrix}
            1 & 1 \\
            0 & 1
        \end{pmatrix}
        =
    \begin{pmatrix}
        \lambda & 0 \\
        0 & 1
    \end{pmatrix}
    \begin{pmatrix}
        1 & \frac{1}{\lambda} \\
        0 & 1
    \end{pmatrix}
    \begin{pmatrix}
        \frac{1}{\lambda} & 0 \\
        0 & 1
    \end{pmatrix}
    \]
    So $f$ conjugate to $g$ where $g(z) = z + 1$.
\end{enumerate}

\setcounter{customtheorem}{16}
\begin{theorem}
    Any non-identity M\"obius map is conjugate to
    one of
    \begin{enumerate}[(i)]
        \item $z \mapsto \nu z$, $\nu \neq 0, 1$
        \item $z \mapsto z + 1$.
    \end{enumerate}
\end{theorem}

\setcounter{customcorollary}{7}
\begin{corollary}
    A non-identity M\"obius map $f$ has either
    \begin{enumerate}[(i)]
        \item 2 fixed points or
        \item 1 fixed point.
    \end{enumerate}
\end{corollary}

\begin{proof}
    Suppose $gfg^{-1} = h$. Then $\alpha$ is a
    fixed point of $f$ (i.e. $f(\alpha) = \alpha$)
    if and only if $g(\alpha)$ is a fixed point of
    $h$ (i.e. $h(g(\alpha)) = g(\alpha)$). So
    number of fixed points of $f$ is the same as
    the number of fixed points of $h$. By Theorem
    17 either,
    \begin{itemize}
        \item $f$ conjugate to $z \mapsto \nu z$
            which has 2 fixed points: $0$,
            $\infty$.
        \item or $f$ conjugate to $z \mapsto z +
            1$ which has 1 fixed points; $\infty$.
    \end{itemize}
\end{proof}

\subsection{Circles in $\CC_\infty$}

A Euclidean circle is the set of points in $\CC$
given by some equation
\[ |z - z_0| = r, \qquad r > 0 .\]
A Euclidean line is the set of points in $\CC$
given by some equation
\[ |z - a| = |z - b| \]
A \emph{circle in $\CC_\infty$} is either a
Euclidean circle or a set $L \cup \{\infty\}$
where $L$ is a Euclidean line. Its general
equation is of the form
\[ Az\ol{z} + B\ol{z} + \ol{B}z + C = 0 \]
for some $A, C \in \RR$, $|B|^2 > AC$. Where $z =
\infty$ is a solution if and only if $A = 0$.
\begin{itemize}
    \item $A = 0$: line
    \item $C = 0$: goes through origin
\end{itemize}
There is a unique circle passing through any 3
distinct points in $\CC_\infty$.

\begin{theorem*}
    Let $f \in \mathcal{M}$ and $C$ a circle in
    $\CC_\infty$, then $f(C)$ is a circle in
    $\CC_\infty$.
\end{theorem*}

\begin{proof}
    By proposition 13, just need to consider $f(z)
    = az$, $z + b$ or $\frac{1}{z}$. Let $S_{A, B,
    C}$ be circle defined by ($*$). Then
    \[ f(z) = az : S_{A, B, C} \mapsto
    S_{A/a\ol{a}, B/\ol{a}, C} \]
    \[ f(z) = z + b : S_{A, B, C} \mapsto S_{A, B
    - Ab, C + Ab\ol{b} - \ol{B}b - B\ol{b}} \]
    \[ f(z) = \frac{1}{z} := \omega : S_{A, B, C}
    \mapsto A + B\omega + B\omega +
    \ol{B}\ol{\omega} + C\omega\ol{\omega} = 0 =
    S_{C, \ol{B}, A} \]
\end{proof}

\bigskip

\noindent
e.g. Consider the image of $\RR \cup \{\infty\}$
under
\[ f(z) = \frac{z - i}{z + i} .\]
It is a circle in $\CC_\infty$ containing
\[ f(0) = -1, f(\infty) = 1, f(1) = -i \]
So $f(\RR \cup \{\infty\}) = \text{unit circle}$.
Furthermore, complimentary components are mapped
to complimentary components.
\begin{center}
\begin{tsqx}
    (-1.5,0)--(1.5,0)--(1.5,1.5)--(-1.5,1.5)--cycle lightblue 0.5 / lightblue
    (-1.5,0)--(1.5,0)
    (0,-1.5)--(0,1.5)
    label $i$ @ (-0.1,0.5)
    ! dot((0,0.5));
    CR (5,0) 0.5 lightblue 0.5 / black
    label $i$ @ (5.1,0.1)
    (3.5,0)--(6.5,0)
    (5,-1.5)--(5,1.5)
    label $\rightarrow$ @ (2.5,0)
\end{tsqx}
\end{center}

\subsection{Cross-Ratios}

\setcounter{customdefinition}{22}
\begin{definition}
    The cross-ratio of distinct points $z_1, z_2,
    z_3, z_4 \in \CC$ is defined by
    \[ [z_1, z_2, z_3, z_4] = \frac{(z_1 -
    z_3)(z_2 - z_4)}{(z_1 - z_2)(z_3 - z_4)} \]
    \[ [\infty, z_2, z_3, z_4] = \frac{(z_2 -
    z_4}{(z_3 - z_4)} \]
    \[ [z_1, \infty z_3, z_4] = - \frac{(z_1 -
    z_3)}{(z_3 - z_4)} \]
    \[ [z_1, z_2, z_3, \infty] = \frac{(z_1 -
    z_3)}{(z_1 - z_2)} \]
    \[ [z_1, z_2, \infty, z_4] = - \frac{(z_2 -
    z_4)}{(z_1 - z_2)} \]
\end{definition}
\noindent
Note $[0, 1, \omega, \infty] = \omega$.

\begin{notation*}
    Different authors use different permutations
    of 1, 2, 3, 4 as definition.
\end{notation*}

\begin{theorem*}
    Given $z_1, z_2, z_3, z_4 \in \CC_\infty$
    distinct and $\omega_1, \omega_2, \omega_3,
    \omega_4 \in \CC_\infty$ distinct then there
    exists $f \in \mathcal{M}$ such that $f(z_i) =
    f(\omega_i)$ if and only if
    \[ [z_1, z_2, z_3, z_4] = [\omega_1, \omega_2,
    \omega_3, \omega_4] .\]
    In particular, M\"obius maps preserve
    cross-ratios
    \[ [z_1, z_2, z_3, z_4] = [f(z_1), f(z_2),
    f(z_3), f(z_4)] .\]
\end{theorem*}

\begin{proof}
    For the forward direction, suppose $f(z_j) =
    \omega_j$ and $z_i, \omega_i \neq \infty$ for
    all $i$ and
    \[ f(z) = \frac{az + b}{cz + d} \]
    then $cz_j + d \neq 0 \forall j$. So
    \begin{align*}
        \omega_j - \omega_k
        &= f(z_j) - f(z_k) \\
        &+ \frac{(ad - bc)(z_j - z_k)}{(cz_j +
        d)(cz_k + d)} \\
        \implies [z_1, z_2, z_3, z_4]
        &= [\omega_1, \omega_2, \omega_3,
        \omega_4] \\
        &= [f(z_1), f(z_2), f(z_3), f(z_4)]
    \end{align*}
    Need to check other cases; $z_i = \infty$,
    $\omega_i = f(\infty = \frac{a}{c}$ etc. \\
    For the other direction, suppose that
    \[ [z_1, z_2, z_3, z_4] = [\omega_1, \omega_2,
    \omega_3, \omega_4] \]
    Let $g \in \mathcal{M}$ such that $g(z_1) =
    0$, $g(z_2) = 1$ and $g(z_4) = \infty$. Let $h
    \in \mathcal{M}$ such that $h(\omega_1) = 0$,
    $h(\omega_2) = 1$, $h(\omega_4) = \infty$.
    Then
    \begin{align*}
        g(z_3)
        &= [0, 1, g(z_3), \infty] \\
        &= [g(z_1), g(z_2), g(z_3), g(z_4)] \\
        &= [z_1, z_2, z_3, z_4] \\
        &= [\omega_1, \omega_2, \omega_3,
        \omega_4] \\
        &= [h(\omega_1), h(\omega_2), h(\omega_3),
        h(\omega_4)] \\
        &= [0, 1, h(\omega), \infty]
        &= h(\omega_3)
    \end{align*}
    So $h^{-1}g$ is the required map.
\end{proof}

\noindent
So $[z_1, z_2, z_3, z_4] = f(z_3)$ where $f$ is
the unique M\"obius map that sends $z_1 \mapsto
0$, $z_2 \mapsto 1$, $z_4 \mapsto \infty$.

\begin{corollary*}
    $z_1, z_2, z_3, z_4$ lie in some circle in
    $\CC_\infty$ if and only if $[z_1, z_2, z_3,
    z_4] \in \RR$.
\end{corollary*}

\begin{proof}
    $C$ circle through $z_1, z_2, z_4$,
    Let $g : C \to \RR \cup \{\infty\}$,
    \[ g(z_1) = 0, g(z_2) = 1, g(z_4) = \infty \]
    \begin{align*}
        g(z_3)
        &= [0, 1, g(z_3), \infty] \\
        &= [g(z_1), g(z_2), g(z_3), g(z_4)] \\
        &= [z_1, z_2, z_3, z_4]
    \end{align*}
    By Theorem 19. So
    \[ [z_1, z_2, z_3, z_4] \in \RR \iff g(z_3)
    \in \RR \iff z_3 \in C .\]
\end{proof}

\begin{center}
    \bfseries\LARGE
    THE END
\end{center}

\end{document}
