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\subsection{Degree of irreducible representations}

\begin{flashcard}[dimV-divides-size-G]
\begin{theorem*}
  If $V$ is a \gls{simple} \gls{rep} of $G$ then
  $\dim V \mid |G|$.
\end{theorem*}

\begin{proof}
  \cloze{Let $\chichar = \chichar_V$ and we'll show
  $\frac{|G|}{\dim V} \in \algO \cap \QQ = \ZZ$.
  \begin{align*}
    \frac{|G|}{\dim V}
    &= \frac{1}{\chichar(e)} \sum_{g \in G}
    \chichar(g) \chichar(g^{-1}) \\
    &= \frac{1}{\chichar(e)} \sum_{i = 1}^r
    |[g_i]_G| \chichar(g_i) \chichar(g_i^{-1}) \\
    &= \sum_{i = 1}^r \homega_{\chichar}(g_i)
    \chichar(g_i^{-1}) \\
    &\in \algO
  \end{align*}
  since $\algO$ is a ring and $\chichar(g_i^{-1})$
  and $\homega_{\chichar}(g_i)$ are all in
  $\algO$. But also $\frac{|G|}{\dim V} \in \QQ$,
  so $\frac{|G|}{\dim V} \in \ZZ = \algO \cap \QQ$
  as required.}
\end{proof}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $G$ is a $p$-group and $\chichar$ is
      an \gls{irred} \gls{char_rep} then
      $\chichar(e)$ is always a power of $p$. In
      particular, if $|G| = p^2$ then as
      $\sum_{\chichar \in \Irr(G)} \chichar(e)^2 =
      |G|$ we see that $\chichar(e) = 1$ for all
      $\chichar \in \Irr(G)$, i.e. $G$ is abelian.
    \item If $G$ is $A_n$ or $S_n$ and $p > n$ is
      prime, then $p$ can't divide the
      \gls{rep_deg} of an \gls{irred} \gls{rep}.
  \end{enumerate}
\end{example*}
\vspace{-1em}

In fact a strange result is true.

\begin{flashcard}[burnside-1904]
\begin{theorem*}[Burnside (1904)]
  \cloze{If $(\rho, V)$ is a \gls{simple} \gls{rep} of
  $G$ then we have $\dim V ~\big|~ |G / Z(G)|$.}
\end{theorem*}
\vspace{-1em}

\cloze{Compare to $|[g]_G| = \frac{|G|}{|C_G(g)|}
~\bigg|~
\frac{|G|}{|Z(G)|}$ for all $g \in G$.}

\begin{proof}
  \cloze{If $Z = Z(G)$ then by \nameref{schurs_lemma}
  $\rho|_Z : Z \to \GL(V)$ has image contained in
  $k^* \id_V$. $\rho(z) = \lambda_z \id_V$ say for
  each $z \in Z$. For each $m \ge 2$ consider the
  \gls{irred} \gls{rep} of $G^m = \ub{G \times G
  \times
  \cdots \times G}_{\text{$m$ times}}$ given by
  $\tpower \rho^m : G^m \to \GL(\tpower V^m)$. If
  $z = (z_1, \ldots, z_m) \in Z^m$ then
  \begin{align*}
    \tpower \rho^m(z)
    &= \sum_{i = 1}^m \lambda_{z_i} \id_{\tpower
    V^m} \\
    &= \lambda_{\left( \prod_{i = 1}^m z_i
    \right)} \id_{\tpower V^m}
  \end{align*}
  So if $\prod_{i = 1}^m z_i = 1$ then $z \in \ker
  \tpower \rho^m$. So $\tpower V^m$ can be viewed
  as an \gls{irred} \gls{rep_deg} $(\dim V)^m$
  \gls{rep} of $\frac{G^m}{Z'}$ where
  \[ Z' = \left\{(z_1, \ldots, z_m) \in Z^m
  ~\bigg|~ \prod_{i = 1}^m z_i = 1 \right\} \le Z^m .\]
  Moreover $|Z'| = |Z|^{m - 1}$. So by previous
  theorem $(\dim V)^m \mid \frac{|G|^m}{|Z|^{m -
  1}}$. Now if $p$ is a prime and $p^a \mid \dim
  V$ then $p^{am} \mid \frac{|G|^m}{|Z|^{m - 1}} =
  \left| \frac{G}{Z} \right|^m |Z|$. By taking $m$
  large enough that $p^m \nmid |Z|$, we see that
  $p^a \mid \left| \frac{G}{|} \right|$. Thus
  $\dim V \mid \left| \frac{G}{Z} \right|$ as
  claimed.}
\end{proof}
\end{flashcard}

\begin{flashcard}[simple-has-no-deg-2-irred-rep]
\begin{proposition*}
  If $G$ is a simple group then $G$ has no
  \gls{irred} \glspl{rep} of \gls{rep_deg} $2$.
\end{proposition*}

\begin{proof}
  \cloze{
  If $G$ is abelian then all \gls{irred}
  \glspl{rep} have \gls{rep_deg} $1$. So we may
  assume that $G$ is non-abelian. If $|G|$ is
  even then $\exists x \in G$ of order $2$. By
  \es[2]{2}, if $\chichar$ is an \gls{irred}
  \gls{char_rep} of $G$ then $\chichar(x) \equiv
  \chichar(e) \pmod{4}$. So if $\chichar(e) = 2$
  then $\chichar(x) = \pm 2$ so $\rho(x) = \pm I$.
  Thus $\rho(x) \in Z(\rho(G))$, \contradiction.
  ($G$ is non-abelian and simple, and $\rho$ is
  non-trivial).

  Now if $|G|$ is odd, we're done by (either of)
  today's theorems so far.}
\end{proof}
\end{flashcard}

\subsection{Burnside's $p^a q^b$ theorem}
\label{sub7_4}

\begin{lemma*}
  Suppose $0 \neq \alpha \in \algO$ is of the form
  $\frac{1}{m} \sum_{i = 1}^m \lambda_i$ for some
  $\lambda_i \in \CC$ such that $\lambda_i^n =
  1$ for some $n \in \NN$. Then $|\alpha| = 1$
  (and so all $\lambda_i$ are equal).
\end{lemma*}

\begin{proof}[Sketch-proof (non-examinable)]
  See \courseref[Galois Theory]{Galois} for the
  details.

  By assumption, $\alpha \in \QQ(\eps)$, $\eps =
  e^{2\pi i/n}$. Let $\mathcal{G} =
  \Gal(\QQ(\eps) / \QQ)$. It is known that
  \[ \{\beta \in \QQ(\eps) \mid \sigma(\beta) =
  \beta ~\forall \sigma \in \mathcal{G}\} = \QQ .\]
  Consider $N(\alpha) = \prod_{\sigma \in
  \mathcal{G}} \sigma(\alpha)$. It is easy to
  verify that $\sigma(N(\alpha)) = N(\alpha)$ for
  all $\sigma \in \mathcal{G}$, i.e. $N(\alpha)
  \in \QQ$. Moreover, $N(\alpha) \in \algO$ since
  if $\alpha$ satisfies a monic integer polynomial
  then every $\sigma(\alpha)$ ($\sigma \in
  \mathcal{G}$) satisfies the same polynomial.
  Thus $N(\alpha) \in \ZZ$. But for each $\sigma
  \in \mathcal{G}$,
  \[ |\sigma(\alpha)| = \left| \frac{1}{m} \sum_{i
  = 1}^m \sigma(\lambda_i) \right|\le 1 .\]
  So $N(\alpha) = \pm 1$ and $|\alpha| = 1$.
\end{proof}

\begin{flashcard}[coprime-char-conj-size-implies-0-or-e]
\begin{lemma*}
  Suppose $\chichar$ is an \gls{irred}
  \gls{char_rep} of $G$ and $g \in G$ such that
  $\chichar(e)$ and $|[g]_G|$ are coprime. Then
  $|\chichar(g)| = \chichar(e)$ or $|\chichar(g)|
  = 0$.
\end{lemma*}
\vspace{-1em}

Note that if $|\chichar(g)| = \chichar(e)$ then
$g$ acts as a scalar on the corresponding
\gls{rep} $V$ and so $\rho(g) \in Z(\rho(G))$.

\begin{proof}
  \cloze{
  By Bezout's lemma, we can find $a, b \in \ZZ$
  such that $a\chichar(e) + b|[g]_G| = 1$. Then
  \[ a\chichar(g) + b \left( \frac{|[g]_G|
  \chichar(g)}{\chichar(e)} \right) =
  \frac{\chichar(g)}{\chichar(e)} \eqdef \alpha \in \algO \]
  Since $\chichar(g)$ is a sum of $\chichar(e)$
  $|G|$-th roots of unity, it follows from the
  last lemma that $\alpha = 0$ or $|\alpha| = 1$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[finite-non-abel-simple-no-prime-power-conj-class]
\begin{proposition*}
  If $G$ is a finite non-abelian group with $g
  \neq e$ such that $|[g]_G|$ has prime power
  order, then $G$ is not simple.
\end{proposition*}

\begin{proof}
  \cloze{Suppose for contradiction that $G$ is simple and
  $g \in G \setminus \{e\}$ such that $|[g]_G| =
  p^r$ for some prime $p$. If $\chichar \in
  \Irr(G) \setminus \{\indicator{G}\}$ then
  $|\chichar(g)| < \chichar(e)$ since otherwise
  $\rho(g)$ is a scalar and lies in
  $Z(\rho(g))  = 1$. Thus by the last lemma, for
  every non-trivial character $\chichar$, either
  $p \mid \chichar(e)$ or $\chichar(g) = 0$. By
  column orthogonality,
  \[ 0 = \sum_{\chichar \in \Irr(G)} \chichar(e)
  \chichar(g)\]
  Thus
  \[ -\frac{1}{p} = \sum_{\substack{\chichar \in
  \Irr(G) \\ \chichar \neq \indicator{}}}
  \frac{\chichar(e)}{p} \chichar(g) \in \algO \cap
  \QQ = \ZZ \quad \contradiction \qedhere \]}
\end{proof}
\end{flashcard}

\begin{flashcard}[burnside-pa-qb-thm]
\begin{theorem*}[Burnside 1904]
  Let $p, q$ be primes and $G$ a group of order
  $p^a q^b$ with $a, b \ge 0$ and $a + b \ge 2$.
  Then $G$ is not simple.
\end{theorem*}

\begin{proof}
  \cloze{Without loss of generality $b > 0$. Let $\QQ$ be
  a Sylow-$p$-subgroup of $G$ and pick $g \in
  Z(\QQ) \setminus \{e\}$ (possible since $\QQ$ is
  a $q$-group). Now $q^b \mid |C_G(g)|$ so
  $|[g]_G| = p^r$ for some $0 \le r \le a$. The
  Theorem follows from the last proposition.}
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item It follows that every group of order
      $p^a q^b$ is soluble, i.e. there exists a
      chain of subgroups $G = G_0 \ge G_1 \ge G_2
      \ge \cdots \ge G_n = \{e\}$ such that for
      all $i$, $G_{i + 1} \normalsub G_i$ and
      $G_i / G_{i + 1}$ is abelian.
    \item Note that $|A_5| = 2^2 \cdot 3 \cdot 5$
      so a finite simple group can have precisely $3$ prime
      factors. Conjugacy classes are $1, 15, 20,
      12, 12$ not prime power order.
    \item The first purely group theoretic proof
      of the $p^a q^b$-theorem first appeared in
      1972.
  \end{enumerate}
\end{remark*}