% vim: tw=50
% 26/01/2023 11AM

\begin{example*}
    Let $G$ act on itself by conjugation, i.e. $g * x = gxg^{-1}$.
\end{example*}

\begin{flashcard}[conj-class]
\begin{definition*}
    $\orb_G(x) = \cloze{\{gxg^{-1} \mid g \in G\}} =
    \cloze{\ccl_G(x)}$ -- the conjugacy class of $x$ in
    $G$.
\end{definition*}
\end{flashcard}

\begin{flashcard}[centraliser]
\begin{definition*}
    $G_x = \cloze{\{g \in G \mid gx = xg\}} =
    \cloze{C_G(x)} \le
    G$ -- the centraliser of $x$ in $G$.
\end{definition*}
\end{flashcard}

\begin{definition*}
    $\ker(\phi) = \{g \in G \mid gx = xg, \forall
    x \in G\} = Z(G)$ -- center of $G$.
\end{definition*}

\begin{note*}
    The map $\phi(g) \colon G \to G$, $h \mapsto
    ghg^{-1}$ satisfies
    \begin{align*}
        \phi(g)(h_1h_2)
        &= gh_1h_2g^{-1} \\
        &= gh_1g^{-1} gh_2 g^{-1} \\
        &= \phi(g)(h_1) \phi(g)(h_2)
    \end{align*}
    so $\phi(g)$ is a group homomorphism, and also a
    bijection, so $\phi(g)$ is an isomorphism.
\end{note*}

\begin{definition*}
    \[ \Aut(G) = \{\text{group isomorphism } f
    \colon G \to G\} \]
\end{definition*}

\noindent
Then $\Aut(G) \le \Sym(X)$ and $\phi \colon G \to
\Sym(X)$ has image in $\Aut(G)$.

\begin{example*}
    Let $X$ be the set of all subgroups of $G$.
    Then $G$ acts on $X$ by conjugation, i.e. $g *
    H = gHg^{-1}$. The stabiliser of $H$ is 
    \[ \{g \in G \mid gHg^{-1} = H\} = N_G(H) \]
    the \emph{normaliser} of $H$ in $G$. This is
    the largest subgroup of $G$ containing $H$ as
    a normal subgroup.
\end{example*}

\begin{hiddenflashcard}[normaliser]
\begin{definition*}[Normaliser]
    The \emph{normaliser} of a subgroup $H$ of $G$
    is
    \[ N_G(H) = \cloze{\{g \in G \mid gHg^{-1} =
    H\}} \]
\end{definition*}
\end{hiddenflashcard}

\newpage
\section{Alternating Groups}

Part IA: elements in $S_n$ are conjugate if and
only if they have the same cycle type.

\begin{example*}
    In $S_5$, we have
    \begin{center}
        \begin{tabular}{c|c}
            \text{cycle type} & \text{\# elements} \\
            \hline
            $\id$ & $1$ \\
            $(*~*)$ & $10$ \\
            $(*~*)(*~*)$ & $15$ \\
            $(*~*~*)$ & $20$ \\
            $(*~*~*)(*~*)$ & $20$ \\
            $(*~*~*~*)$ & $30$ \\
            $(*~*~*~*~*)$ & $24$ \\
            \hline
            \text{total} & $120$
        \end{tabular}
    \end{center}
\end{example*}

\noindent
Let $g \in A_n$. Then $C_{A_n}(g) = C_{S_n}(g)
\cap A_n$ if there exists odd permutation
commuting with $g$. Then $|C_{A_n}(g)| = \half
|C_{S_n}(g)|$ and $|\ccl_{A_n}(g)| =
|\ccl_{S_n}(g)|$ otherwise $|C_{A_n}(g)| =
|C_{S_n}(g)|$ and $|\ccl_{A_n}(g)| = \half
|\ccl_{S_n}(g)|$.

\begin{example*}
    Taking $n = 5$, $(1~2)(3~4)$ commutes with
    $(1~2)$ and $(1~2~3)$ commutes with $(4~5)$ (and
    $(1~2)$ and $(4~5)$ are both odd). But if $h \in
    C_{S_5}(g)$ where $g = (1~2~3~4~5)$, then
    $(1~2~3~4~5) = h(1~2~3~4~5)h^{-1} =
    (h(1)~h(2)~h(3)~h(4)~h(5))$. So $h \in \langle
    g \rangle \le A_5$. $|\ccl_{A_5}(g)| = \half
    |\ccl_{A_5}(g)| = 12$. Thus $A_5$ has
    conjugacy classes of sizes $1, 15, 20, 12,
    12$.

    \myskip
    If $H \normalsub A_5$, then $H$ is a
    union of conjugacy classes.  So $|H| = 1 + 15a
    + 20b + 12c$ for some integers $a, b \in \{0,
    1\}$, $c \in \{0, 1, 2\}$ and by Lagrange's
    Theorem $|H| \big| 60$. One can check that
    the only way that this can happen is if $|H| =
    1$ or $|H| = 60$. So $A_5$ is simple.
\end{example*}

\begin{flashcard}[alternating-group-generated]
\begin{lemma}
    $A_n$ is generated by \cloze{3-cycles.}
\end{lemma}
\end{flashcard}

\begin{proof}
    Each $\sigma \in A_n$ is product of an even
    number of transpositions. Thus suffices to
    write the product of any two transpositions as
    a product of 3-cycles.

    \myskip
    For $a, b, c, d$ distinct, the possible
    distinct cases are $(a~b)(a~b)$, $(a~b)(b~c)$
    and $(a~b)(c~d)$. We can check these are all a
    product of 3-cycles:
    \begin{align*}
        (a~b)(a~b)
        &= \id \\
        (a~b)(b~c)
        &= (a~b~c) \\
        (a~b)(c~d)
        &= (a~c~b)(a~c~d)
        \qedhere
    \end{align*}
\end{proof}

\begin{flashcard}[alternating-conjugate-three-cycles]
\begin{lemma}
    If \cloze{$n \ge 5$} then all 3-cycles in $A_n$ are
    conjugate.
\end{lemma}
\end{flashcard}

\begin{proof}
    We claim that any 3-cycle is conjugate to
    $(1~2~3)$. Indeed if $(a~b~c)$ is a 3-cycle
    then $(a~b~c) = \sigma(1~2~3) \sigma^{-1}$ for
    some $\sigma \in S_n$. If $\sigma \not\in A_n$
    then replace by $\tilde{\sigma} = \sigma
    (4~5)$.
\end{proof}

\begin{theorem}
    $A_n$ is simple for all $n \ge 5$.
\end{theorem}

\begin{proof}
    Let $1 \ne N \normalsub A_n$. Suffices to show
    that $N$ contains a 3-cycle, since by Lemma
    3.1 and Lemma 3.2 we have $N = A_n$.

    \myskip
    Take $1 \neq \sigma \in N$ and write $\sigma$
    as a product of disjoint cycles.
    \begin{itemize}
        \item Case 1: $\sigma$ contains a cycle of
            length $r \ge 4$. Without loss of
            generality $\sigma = (1~2\cdots
            r)\tau$. Let $\delta = (1~2~3)$. Then
            \begin{align*}
                \ub{\sigma^{-1}}_{\in N}
                \ub{\delta^{-1}\sigma\delta}_{\in
                N}
                &= (r\cdots 2~1)(1~3~2)
                (1~2~3\cdots r)(1~2~3) \\
                &= (2~3~r)
            \end{align*}
            So $N$ contains a 3-cycle.
        \item Case 2: $\sigma$ contains two
            3-cycles. Without loss of generality
            $\sigma = (1~2~3)(4~5~6) \tau$. Let
            $\delta = (1~2~4)$. Then
            \begin{align*}
                \ub{\sigma^{-1}}_{\in N}
                \ub{\delta^{-1}\sigma\delta}_{\in
                N}
                &= (1~3~2)(4~6~5)(1~4~2)
                (1~2~3)(4~5~6)(1~2~4) \\
                &= (1~2~4~3~6)
            \end{align*}
            So now done by case 1.
        \item Case 3: $\sigma$ contains two
            2-cycles. Without loss of generality
            $\sigma = (1~2)(3~4) \tau$. Let
            $\delta = (1~2~3)$. Then
            \begin{align*}
                \ub{\sigma^{-1}}_{\in N}
                \ub{\delta^{-1} \sigma\delta}_{\in
                N}
                &= (1~2)(3~4)(1~3~2)(1~2)(3~4)
                (1~2~3) \\
                &= (1~4)(2~4)
            \end{align*}
            Let $\eps = (2~3~5)$ ($n \ge 5$). Then
            \begin{align*}
                \ub{\pi^{-1}\eps^{-1}\pi
                \eps}_{\in N}
                &= (1~4)(2~3)(2~5~3)(1~4)(2~3)
                (2~3~5) \\
                &= (2~5~3)
            \end{align*}
            So $N$ contains a 3-cycle.
    \end{itemize}
    Conclusion of proof: Remains to consider
    $\sigma$ with one of these cycle types:
    \begin{itemize}
        \item Case $(*~*)$ or $(*~*)(*~*~*)$ but
            then $\sigma \not\in A_n$,
            contradiction.
        \item Case $(*~*~*)$ but then $\sigma$ is
            a 3-cycle so we're already done.
            \qedhere
    \end{itemize}
\end{proof}