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\newpage
\section{Group Actions}

\begin{definition*}
    For $X$ a set, let $\Sym(X)$ be the group of
    all bijections $X \to X$ under composition
    (identity $\id = \id_X$).
\end{definition*}

\begin{definition*}
    A group $G$ is a permutation group of degree
    $n$ if $G \le \Sym(X)$ with $|X| = n$.
\end{definition*}

\begin{example*}
    $S_n = \Sym(\{1, 2, \ldots, n\})$ is a
    permutation group of degree $n$, as is $A_n
    \le S_n$. $D_{2n} = \{\text{symmetries of a
    regular $n$-gon}\}$ so is a subgroup of $S_n
    \cong \Sym(\{\text{vertices of $n$-gon}\})$.
\end{example*}

\begin{definition*}
    An action of a group $G$ on a set $X$ is a
    function $* \colon G \times X \to X$
    satisfying
    \begin{enumerate}[(i)]
        \item $e * x = x$ for all $x \in X$
        \item $(g_1 g_2) * x = g_1 * (g_2 * x)$
            for all $g_1, g_2 \in G$ and for all
            $x \in X$.
    \end{enumerate}
\end{definition*}

\begin{proposition}
    An action of a group $G$ on a set $X$ is
    equivalent to specifying a group homomorphism
    $\phi \colon G \to \Sym(X)$.
\end{proposition}

\begin{proof}
    For each $g \in G$, let $\phi_g \colon X \to
    X$, $x \mapsto g * x$. We have
    \begin{align*}
        \phi_{g_1 g_2}(x)
        &= (g_1 g_2) * x \\
        &= g_1 * (g_2 * x) \\
        &= \phi_{g_1}(g_2 * x) \\
        &= \phi_{g_1} \circ \phi_{g_2}(x)
    \end{align*}
    Then $\phi_{g_1 g_2} = \phi_{g_1} \circ
    \phi_{g_2}$ (\dag).

    \myskip
    In particular, $\phi_g \circ \phi_{g^{-1}} =
    \phi_{g^{-1}} \circ \phi_g = \phi_e = \id$.
    Thus $\phi_y \in \Sym(X)$.

    \myskip
    Define $\phi \colon G \to \Sym(X)$, $g \mapsto
    \phi_g$ (a group homomorphism by (\dag)).
    Conversely let $\phi \colon G \to \Sym(X)$ be
    a group homomorphism. Define $* \colon G
    \times X \to X$, $(g, x) \mapsto \phi(g)(x)$.
    Then
    \begin{enumerate}[(i)]
        \item $e * x = \phi(e)(x) = \id(x) = x$.
        \item \begin{align*}
                (g_1 g_2) * x
                &= \phi(g_1 g_2)(x) \\
                &= \phi(g_1) \circ \phi(g_2)(x) \\
                &= g_1 * (g_2 * x)
                \qedhere
            \end{align*}
    \end{enumerate}
\end{proof}

\begin{definition*}
    We say $\phi \colon G \to \Sym(X)$ is a
    permutation representation of $G$.
\end{definition*}

\begin{definition*}
    Let $G$ act on a set $X$.
    \begin{enumerate}[(i)]
        \item The orbit of $x \in X$ is
            \[ \orb_G(x) = \{g \in x \mid g \in
            G\} \subseteq X .\]
        \item The stabiliser $x \in X$ is
            \[ G_x = \{g \in G \mid g * x = x\}
            \le G .\]
    \end{enumerate}
\end{definition*}

\noindent
Recall Groups IA: Orbit-Stabiliser theorem. There
is a bijection
\[ \orb_G(x) \leftrightarrow G / G_x \]
(where $G / G_x$ is the set of left cosets of
$G_x$ in $G$). In particular if $G$ is finite,
\[ |G| = |\orb_G(x)| |G_x| \]

\begin{example*}
    Let $G$ be the group of all symmetries of a
    cube. $X = \text{set of vertices}$, $x \in X$,
    $|\orb_G(x)| = 8$, $|G_x| = 6$.
    \begin{center}
        \includegraphics[width=0.2\linewidth]
        {images/434843409be311ed.png}
    \end{center}
    Hence $|G| = 48$.
\end{example*}

\begin{remark*}
    \begin{enumerate}[(i)]
        \item $\ker \phi = \bigcap_{x \in X} G_x$
            is called the kernel of the group
            action.
        \item The orbits partition $X$. We say the
            action is \emph{transitive} if there
            is only one orbit.
        \item $G_{g * x} = g G_x g^{-1}$, so if
            $x, y \in X$ belong to the same orbit,
            then their stabilizers are conjugate.
    \end{enumerate}
\end{remark*}

\subsubsection*{Examples}

\begin{example*}
    Let $G$ act on itself by left multiplication,
    i.e. $g * x = g \cdot x$. The kernel of this action
    is
    \[ \{g \in G \mid g \cdot x = x ~\forall x \in
    G\} = \{e\} \]
    Thus $G \injto \Sym(G)$. This proves:
    \begin{theorem}[Cayley's Theorem]
        Any finite group $G$ is isomorphic to a
        subgroup of $S_n$ for some $n$. (Take $n =
        |G|$).
    \end{theorem}
\end{example*}

\begin{example*}
    Let $H \le G$. $G$ acts on $G / H$ (left
    cosets) by left multiplication, i.e. $g * xH =
    gxH$. This action is transitive (since $(x_2
    x_1^{-1}) x_1 H = x_2 H$) with
    \[ G_{xH} = \{g \in G \mid gxH = xH\} = \{g
    \in G \mid x^{-1} gx \in H\} = x^{-1}Hx \]
    Thus $\ker(\phi) = \bigcap_{x \in G}
    xHx^{-1}$. This is largest normal subgroup of
    $G$ that is contained in $H$.
\end{example*}

\begin{theorem}
    let $G$ be a non-abelian simple group, and $H
    \le G$ a subgroup of index $n > 1$. Then $n \ge
    5$ and $G$ is isomorphic to a subgroup of
    $A_n$.
\end{theorem}

\begin{proof}
    Let $G$ act on $X = G / H$ by left
    multiplication and let $\phi \colon G \to
    \Sym(X) = S_n$ be the associated permutation
    representation. As $G$ is simple, $\ker(\phi)
    = 1$ or $\ker(\phi) = G$. If $\ker(\phi) = G$,
    then $\Im(\phi) = 1$, contradiction since $G$
    acts transitively on $X$ and $|X| > 1$. Thus
    $\ker(\phi) = 1$ and $G \cong \Im(\phi) \le
    S_n$. Since $G \le S_n$ and $A_n \normalsub
    S_n$, second isomorphism theorem gives:
    \[ G \cap A_n \normalsub G \]
    and
    \[ G / G \cap A_n \cong G A_n / A_n \le S_n /
    A_n \cong C_2 \]
    $G$ simple implies that $G \cap A_n = 1$ or
    $G$. If it equals $1$ then $G \injto C_2$
    contradicts $G$ non-abelian. If it equals $G$
    then $G \le A_n$. Finally, if $n \le 4$, then
    $A_n$ has no non-abelian simple subgroup (just
    list them!).
\end{proof}