% vim: tw=50
% 02/02/2023 12AM

\begin{example}
    Let $G = \PSL_2(\ZZ / 5\ZZ)$. Then $|G| =
    \frac{4 \times 5 \times 6}{2} = 60 = 2^2
    \times 3 \times 5$. Let $G$ act on $\ZZ / 5\ZZ
    \cup \{\infty\}$ via
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    : 2 \mapsto \frac{az + b}{cz + d} \]
    By Lemma 5.2 the permutation representation
    \[ \phi : G \to \Sym(\{0, 1, 2, 3, 4,
    \infty\}) \cong S_6 \]
    is injective. \myskip
    Claim: $\Im(\phi) \le A_6$, i.e. $\psi : G
    \stackrel{\phi}{\longrightarrow} S_6
    \stackrel{\sgn}{\longrightarrow} \{\pm 1\}$ is
    trivial. \\
    Proof: Let $g \in G$ have order $d$. Write $d
    = 2^n m$ with $m$ odd. Then $h^m$ has order
    $2^n$. If $\psi(h^m) = 1$ then $\psi(h)^m = 1$
    so $\psi(h) = 1$. So it suffices to show that
    $\psi(g) = 1$ for all $g \in G$ with order a
    power of 2. \\
    Lemma 4.7 implies every such $g$ belongs to a
    Sylow 2-subgroup. \\
    Therefore it suffices to check $\psi(H) = 1$
    for $H$ a Sylow 2-subgroup. (since $\ker(\psi)
    \normalsub G$ and all Sylow 2-subgroups are
    conjugate). \myskip
    Take
    \[ H = \left\langle
    \begin{pmatrix}
        2 & 0 \\
        0 & 3
    \end{pmatrix}
    \{\pm I\},
    \begin{pmatrix}
        0 & 1 \\
        -1 & 0
    \end{pmatrix}
    \{\pm I\} \right\rangle \le G
    = \frac{\SL_2(\ZZ / 5\ZZ)}{\{\pm I\}} \]
    We compute
    \begin{align*}
        \phi
        \begin{pmatrix}
            2 & 0 \\
            0 & 3
        \end{pmatrix}
        &= (1~4)(2~3) \\
        \phi
        \begin{pmatrix}
            0 & 2 \\
            -1 & 0
        \end{pmatrix}
        &= (0~\infty)(1~4)
    \end{align*}
    Both of these are even, therefore $\psi(H) =
    1$. This proves the claim.
\end{example}

\noindent
On Example Sheet 1, Question 14 we will prove that
if $G \le A_6$ and $|G| = 60$ then $G \cong A_5$.

\subsubsection*{Facts (not proved in this course)}

$\PSL_n(\ZZ / p\ZZ)$ is a simple group $\forall n
\ge 2$, $p$ prime except $(n, p) = (2, 2), (2, 3)$
(these are examples of finite groups of Lie type).
The smallest non-abelian simple groups are
\[ A_5 \cong \PSL_2(\ZZ / 5\ZZ) \]
(order 60) and
\[ \PSL_2(\ZZ / 7\ZZ) \cong \GL_3(\ZZ / 7\ZZ) \]
(order 168).

\newpage
\section{Finite abelian groups}

Later we prove (in the modules chapter)

\begin{theorem}
    Every finite abelian group is isomorphic to a
    product of cyclic groups.
\end{theorem}

\noindent
However it may be possible to write the same group
as a product of cyclic groups in more than one
way.

\begin{lemma}
    If $m, n \in \ZZ_{\ge 1}$ coprime then
    \[ C_m \times C_n \cong C_{mn} \]
\end{lemma}

\begin{proof}
    let $g$ and $h$ be generators of $C_m$ and
    $C_n$. Then $(g, h) \in C_m \times C_n$ and
    $(g, h)^r = (g^r, h^r)$. Then
    \begin{align*}
        (g, h)^r = 1
        &\iff m \mid r \text{ and } n \mid r \\
        &\iff mn \mid r
    \end{align*}
    (since $m, n$ coprime). Thus $(g, h)$ has
    order $mn = |C_m \times C_n|$. Therefore $C_m
    \times C_n \cong C_{mn}$.
\end{proof}

\begin{corollary}
    Let $G$ be a finite abelian group. Then
    \[ G \cong C_{n_1} \times C_{n_2} \times
    \cdots \times C_{n_k} \]
    where each $n_i$ is a prime power.
\end{corollary}

\begin{proof}
    If $n = p_1^{a_1} \cdots p_r^{a_r}$ ($p_1,
    \ldots, p_r$ distinct primes), then Lemma 6.2
    shows
    \[ C_n \cong C_{p_1^a} \times \cdots \times
    C_{p_r^{a_r}} \]
    Writing each of the cyclic groups in Theorem
    6.1 in this way gives the result.
\end{proof}

\myskip
In fact when we prove Theorem 6.1 we will prove
the following refinement:

\begin{flashcard}[characterization-of-finite-abelian-groups]
\begin{theorem}
    Let $G$ be a finite abelian group. Then
    \[ G \cong \cloze{C_{d_1} \times C_{d_2} \times
    \cdots \times C_{d_t}} \]
    for some \cloze{$d_1 \mid D_2 \mid \cdots \mid
    d_t$.}
\end{theorem}
\end{flashcard}

\begin{remark}
    The integers $n_1, \ldots, n_k$ in Corollary
    6.3 (up to ordering) and $d_1, \ldots, d_t$ in
    Theorem 6.4 (assuming $d_1 > 1$) are uniquely
    determined by the group $G$.
\end{remark}

\noindent
(Proof omitted -- but works by counting the number
of elements of $G$ of each prime power order).

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item The abelian groups of order 8 are
        \[ C_8, \quad C_2 \times C_2 \quad
        \text{and} \quad C_2 \times C_2 \times C_2 \]
    \item The abelian groups of order 12 are
        \[ C_2 \times C_2 \times C_3 \cong C_2
        \times C_6 \]
        and
        \[ C_4 \times C_3 \cong C_{12} \]
\end{enumerate}

\begin{flashcard}[exponent-of-a-group-defn]
\begin{definition*}[Exponent of a group]
    \cloze{The \emph{exponent} of a group $G$ is the
    least integer $n \ge 1$ such that $g^n = 1$
    for all $g \in G$, i.e. the lowest common
    multiple of all the orders of the elements of
    $G$.}
\end{definition*}
\end{flashcard}

\begin{example*}
    $A_4$ has exponent 6.
\end{example*}

\begin{corollary}
    Every finite abelian group contains an element
    whose order is the exponent of the group.
\end{corollary}

\begin{proof}
    If $G \cong C_{d_1} \times \cdots C_{d_t}$
    with $d_1 \mid d_2 \mid \cdots \mid d_t$, then
    every $g \in G$ has order dividing $d_t$ and
    if $h \in C_{d_t}$ is a generator then $(1, 1,
    1, \ldots, 1, h) \in G$ has order $d_t$. Thus
    $G$ has exponent $d_t$.
\end{proof}