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\begin{flashcard}[char-proof-of-reg-rep-decomp-thm]
\begin{theorem*}
  If $V_1, \ldots, V_r$ is a complete list of
  \gls{irred} \glspl{rep} of a finite group $G /
  \CC$, then the \gls{reg_rep} $\CC G$ decomposes
  as
  \[ \bigoplus_{i = 1}^r (\dim V_i) V_i .\]
  In particular, $|G| = \sum_{i = 1}^r (\dim
  V_i)^2$.
\end{theorem*}

\begin{proof}
  \cloze{We just need to show
  \[ \dim \GHom_G(\CC G, V_i) = \dim V_i ~\forall
  i \]
  But
  \begin{align*}
    \dim \GHom_G(\CC G, V_i)
    &= \Gip\langle \chichar_{\CC G}, \chichar_{V_i}
    \rangle_G \\
    &= \frac{1}{|G|} \sum_{g \in G} |\{h \in G \st
    gh = h\}| \chichar_{V_i}(g) \\
    &= \frac{1}{|G|} |G| \chichar_{V_i}(e) \\
    &= \chichar_{V_i}(e)
  \end{align*}
  as required.}
\end{proof}
\end{flashcard}

\begin{flashcard}[burnsides-lemma]
\begin{proposition*}[Burnside's Lemma]
  \label{burnsides_lemma}
  Let $G$ be a finite group and $X$ a finite set
  with $G$-action. Then
  \[ \cloze{\Gip\langle \mathbbm{1}, \chichar_{\CC
  X} \rangle_G = \#\text{orbits of $G$ on $X$} .} \]
\end{proposition*}

\begin{proof}
  \cloze{
  \begin{align*}
    |G| \Gip\langle \mathbbm{1}, \chichar_{\CC X}
    \rangle_G
    &= \sum_{g \in G} \chichar_{\CC X}(g) \\
    &= \sum_{g \in G} |\{x \in X \st gx = x\}| \\
    &= |\{(g, x) \in G \times X \st gx = x\}| \\
    &= \sum_{x \in X} |\{g \in G \st gx = x\}|
  \end{align*}
  So
  \begin{align*}
    \Gip\langle \mathbbm{1}, \chichar_{\CC X}
    \rangle_G
    &= \sum_{x \in X} \frac{|\Stab_G(x)|}{|G|} \\
    &= \sum_{x \in X} \frac{1}{|\Orb_G(X)|} \\
    &= \sum_{\text{orbits $O_i$}} \left( \sum_{x
    \in O_i} \frac{1}{|O_i|} \right) \\
    &= \#\text{orbits}
  \end{align*}
  }
\end{proof}
\end{flashcard}

Note that if $X = \bigcup O_i$ is the orbit
decomposition then we've seen before $\CC X =
\bigoplus \CC O_i$, so \nameref{burnsides_lemma}
says each $\CC O_i$ contains precisely one copy of
the \gls{triv_rep} $\CC$ -- the constant functions
on $O_i$. This is not so hard to prove directly
(exercise).

If $X, Y$ are two sets with $G$-actions then $X
\times Y$ is a set with $G$-action via $g \cdot
(x, y) = (g \cdot x, g \cdot y)$ for $(x, y) \in X
\times Y$, $g \in G$.

\begin{lemma*}
  If $X, Y$ are finite, then $\chichar_{\CC X
  \times Y} = \chichar_{\CC X} \cdot \chichar_{\CC
  Y}$.
\end{lemma*}

\begin{proof}
  If $g \in G$,
  \begin{align*}
    \chichar_{\CC X \times Y} (g)
    &= |\{(x, y) \in X \times Y \st (gx, gy) = (x,
    y)\}| \\
    &= |\{x \in X \st gx = x\}| \cdot |\{y \in Y
    \st gy = y\}| \\
    &= \chichar_{\CC X}(g) \chichar_{\CC Y}(g)
    \qedhere
  \end{align*}
\end{proof}

\begin{corollary*}
  If $G$ is a finite group and $X, Y$ are finite
  sets with $G$-actions, then
  \[ \Gip\langle \chichar_{\CC X}, \chichar_{\CC Y}
  \rangle_G = \#\text{$G$-orbits in $X \times Y$} \]
\end{corollary*}

\begin{proof}
  \begin{align*}
    \Gip\langle \chichar_{\CC X}, \chichar_{\CC Y}
    \rangle_G
    &= \frac{1}{|G|} \sum_{g \in G} \chichar_{\CC
    X}(g) \chichar_{\CC Y}(g) \\
    &= \frac{1}{|G|} \sum_{g \in G} 1
    \chichar_{\CC X \times Y} \\
    &= \Gip\langle \mathbbm{1}, \chichar_{\CC X \times
    Y} \rangle_G \\
    &= \#\text{$G$-orbits on $X \times Y$}
    &&\text{\nameref{burnsides_lemma}}
  \end{align*}
\end{proof}

\begin{remark*}
  If $X$ is any set with $G$-action and at least
  $2$-elements then $\{(x, x) \st x \in X\}
  \subset X \times X$ is $G$-stable and non-empty
  and its complement $\{(x, y) \st x, y \in X, x
  \neq \}$ is also non-empty and $G$-stable.
\end{remark*}

\begin{flashcard}[2-transitive-defn]
\begin{definition*}[2-transitive action]
  \glsadjdefn{2trans}{$2$-transitive}{group action}
  \glsadjdefn[2trans]{2transly}{$2$-transitively}{group action}
  \cloze{
  We say $G$ acts \emph{$2$-transitively} on $X$
  if for all $x_1, x_2, y_1, y_2$ with $x_1 \neq
  y_1$, $x_2 \neq y_2$, there exists $g \in G$
  such that $gx_1 = x_2$ and $gy_1 = y_2$ (i.e. $g
  \cdot (x_1, y_1) = (x_2, y_2)$).
  \fcemph{Equivalently} if
  the $G$-action on $X \times X$ has
  \fcemph{precisely} two
  orbits.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  $S_n$ acts \gls{2transly} on $\{1, \ldots, n\}$
  for all $n \ge 2$. If $g$ acts \gls{2transly} on
  $X \times X$ then by the last corollary,
  \[ \Gip\langle \chichar_{\CC X}, \chichar_{\CC X}
  \rangle_G = 2 \]
  So if $\chichar_{\CC X} = \bigoplus_{i = 1}^r n_i
  V_i$, $V_i$ \gls{irred} and pairwise
  non-\gls{rep_iso} then $\sum n_i^2 = 2$. That
  is, $\CC X$ has two non-\gls{rep_iso}
  \gls{irred} summands, namely the constant
  functions and
  \[ V = \{f \in \CC X \st \sum_{x \in X} f(x) =
  0\} \]
  Then $\chichar_V$ is an irreducible
  \gls{char_rep} so
  \[ \chichar_V(g) = \chichar_{\CC X}(g) -
  \indicator{G}(g) = \text{(\# fixed points of $g$
  on $X$)} - 1 \]
  Moreover, if $V$ is \gls{irred} then the action
  must be \gls{2trans}.
\end{example*}
\vspace{-1em}

\textbf{Exercise:} If $G = \GL_2(\FF_p)$ then
decompose the permutation \gls{rep} of $G$ coming
from action of $G$ on $\FF_p \cup \{\infty\}$ by
M\"obius maps:
\[
\begin{pmatrix}
  a & b \\
  c & d
\end{pmatrix}
z = \frac{az + b}{cz + d} \]
if $z \in \FF_p \setminus \{-d/c\}$ etc.

\subsubsection*{Examples}

\begin{enumerate}[(1)]
  \item $G = S_4$. The character table of $S_4$ is
    \begin{center}
      \begin{tabular}{c|c|c|c|c|c}
        $|[g_i]_G|$ & $1$ & $3$ & $8$ & $6$ & $6$
        \\
        \hline
        $g_i$ & $e$ & $(12)(34)$ & $(123)$ &
        $(12)$ & $(1234)$ \\
        \hline
        $\mathbbm{1}$ & $1$ & $1$ & $1$ & $1$ &
        $1$ \\
        $\sum$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
        $\chichar_3$ & $3$ & $-1$ & $0$ & $1$ &
        $-1$ \\
        $\chichar_4$ & $3$ & $-1$ & $0$ & $-1$ &
        $1$ \\
        $\chichar_5$ & $2$ & $2$ & $-1$ & $0$ &
        $0$
      \end{tabular}
    \end{center}
    \begin{proof}
      $\mathbbm{1}, \sum$ are constructed as for
      $S_3$. By our discussion above,
      $\chichar_{\CC\{1, 2, 3, 4\}} = \mathbbm{1}
      + \chichar_V$ for some \gls{irred} \gls{rep}
      $V$ of degree $3$ and we can let $\chichar_3
      = \chichar_V$ such that $\chichar_3(g) =
      \#\text{fixed points of $g$} - 1$. We saw on
      \es[2]{1} that if $\theta$ is a \repdim{1}
      \gls{rep} and $\rho$ is any \gls{irred}
      \gls{rep} then $(\rho \otimes \theta) (g)
      \defeq \theta(g) \rho(g)$ is an \gls{irred}
      \gls{rep} of $G$ and $\chichar_{\rho \otimes
      \theta}(g) = \chichar_\rho(g) \theta(g)$.
      Thus we can set $\chichar_4 = \sum
      \cdot \chichar_3$. We can compute
      $\chichar_5$ via column orthogonality
      \[ 1^2 + 1^2 + 3^2 + 3^2 + \chichar_5(e)^2 =
      24 \]
      so $\chichar_5(e) = 2$, and $\sum_{i = 1}^5
      \chichar_i(e) \chichar_i(e) = 0 ~\forall g
      \in S_4 \setminus \{e\}$.
    \end{proof}
    (In fact there is a homomorphism $S_4 \to S_3$
    giving a \gls{2trans} action of $S_4$ on $\{1,
    2, 3\}$ and $\chichar_5 = \chichar_{\CC\{1, 2,
    3\}} - \mathbbm{1}$).
  \item $G = A_4$. Every \gls{irred} \gls{rep} of
    $S_4$ may be restricted to $A_4$ and its
    \gls{char_rep} values won't change. In this
    way we get $3$ characters of $A_4$
    \[ \mathbbm{1} = \mathbbm{1}_{s_4} |_{A_4},
    \quad \psi_2 = \chichar_3 |_{A_4} = \chichar_3
    |_{A_4}, \quad \psi_3 = \chichar_5|_{A_4} \]
    It is irreducible since  it has dimension 1:
    \[ \langle \psi_2, \psi_2 \rangle_{A_4} =
    \frac{1}{12} (13^2 + 3(-1)^2 + 8\cdot 0^2) = 1 \]
    so $\psi_2$ is irreducible. However,
    \[ \langle \psi_3, \psi_3 \rangle_{A_4} =
    \frac{1}{12} (12^2 + 3(2)^2 + 8(-1)^2) = 2 \]
    so $\psi_3$ decomposes into two
    $1$-dimensional non-\gls{rep_iso} pieces.

    \textbf{Exercise:} Use this to construct the
    character table of $A_4$. Recall
    $[(123)]_{S_4}$ is a union of two classes in
    $A_4$.
\end{enumerate}

\newpage
\section{The Character Ring}

We've already seen that the algebraic structure on
$\clsfn_G$ for a finite group $G$ has \gls{rep}
theoretic meaning, e.g. if $V_1, V_2$ are
\glspl{rep} then
\begin{align*}
  \chichar_{V_1 \oplus V_2}
  &= \chichar_{V_1} + \chichar_{V_2} \\
  \chichar_0
  &= 0 \\
  \chichar_k
  &= \indicator{G} \\
  \Gip\langle \chichar_{V_1}, \chichar_{V_2}
  \rangle_G
  &= \dim \GHom_G(V_1, V_2)
\end{align*}
We've also seen $\chichar_{\CC X \times Y} =
\chichar_{\CC X} \cdot \chichar_{\CC Y}$ and if
$\theta, \rho$ are \glspl{rep} such that $\theta$
is \repdim{1} then $\chichar_{\theta \oplus \rho}
= \chichar_\theta \cdot \chichar_\rho ( = \theta
\chichar_\rho)$. We want to generalise this to any
pair of \glspl{rep} so $\chichar_{\sigma \oplus
\rho} = \chichar_\sigma \cdot \chichar_\rho$.
