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Recall that for $g \in G$,
\[ \FXW(KgH, W)^H = \{f : KgH \to W \st f(xh) =
h^{-1} f(x) ~\forall x \in KgH, h \in H\} \]
is a \gls{rep} of $K$ via
\[ (kf)(x) = f(k^{-1} x) \qquad \forall x \in KgH
\in k \in K .\]
Last time we reduced the proof of
\nameref{mack_rest_form} to the following lemma:

\begin{flashcard}[mackeys-rest-form-sublemma]
\begin{lemma*}
  There is an isomorphism of \glspl{rep} of $K$
  \[ \FXW(KgH, W)^H \simeqto \cloze{\FXW(K, \leftgrep{}^g
  W)^{K \cap \leftgrep{}^g H}} .\]
\end{lemma*}

\begin{proof}
  \cloze{Let $\Theta : \FXW(KgH, W)^H \to \FXW(K,
  \leftgrep{}^g W)$, $\Theta(f)(k) = f(kg)$. If
  $k' \in K$,
  \[ (k' \cdot \Theta(f))(k) = \Theta(f)(k'^{-1}
  k) = f(k'^{-1} g) = (k' \cdot f)(kg) =
  \Theta(k' f)(k) \]
  i.e. $\Theta$ is $k$-linear. If $ghg^{-1} \in K$
  for some $h \in H$, then
  \begin{align*}
    \Theta(f)(kghg^{-1})
    &= f(kgh) \\
    &= \rho(h)^{-1} f(kg) \\
    &= (\leftgrep{}^g \rho)(ghg^{-1}) \Theta(f)(k)
  \end{align*}
  i.e. $\Im\Theta \le \FXW(K, \leftgrep{}^g
  W)^{K \cap \leftgrep{}^g H}$. We try to define
  an inverse to $\Theta$ via
  \begin{align*}
    \psi : \FXW(K, \leftgrep{}^g W)^{K \cap
    \leftgrep{}^g H}
    &\to \FXW(KgH, W)^H \\
    \psi(f)(kgh)
    &= \rho(h^{-1}) f(k)
  \end{align*}
  If $k_1 gh_1 = k_2 gh_2$ then $k_2^{-1} k_1 =
  g(h_2 h_1^{-1}) g^{-1} \in K \cap \leftgrep{}^g
  H$.
  \begin{align*}
    f(k_2)
    &= f(k_1 (k_2^{-1} k_1)^{-1}) \\
    &= (\leftgrep{}^g \rho) (gh_2 h_1^{-1}
    g^{-1})f(k_1) \\
    \rho(h_2 h_1^{-1}) f(k_1)
  \end{align*}
  So $\rho(h_2)^{-1} f(k_2) = \rho(h_1^{-1})
  f(k_1)$ i.e. $\psi(f)$ is well-defined. Moreover
  if $f \in \FXW(KgH, W)^H$, then
  \[ \psi\Theta(f)(kgh) = \rho(h)^{-1}
  \Theta(f)(k) = \rho(h^{-1}) f(kg) = f(kgh) .\]
  and if $f \in \FXW(K, \leftgrep{}^g
  W)^{\leftgrep{}^g H \cap K}$. Also
  \[ \Theta\psi(f)(k) = \psi(f)(kg) = f(k) \]
  so $\psi$ is inverse to $\Theta$.}
\end{proof}
\end{flashcard}

\subsection{Frobenius Groups}

\begin{flashcard}[frobenius-1901]
\begin{theorem*}[Frobenius 1901]
  \label{frob_1901}
  \cloze{Let $G$ be a finite group acting transitively on
  a set $X$. If each $g \in G \setminus \{e\}$
  fixes at most one element of $X$ then
  \[ K = \{e\} \cup \{g \in G \st gx \neq x
  ~\forall x \in X\} \]
  is a normal subgroup of $G$ of order $|X|$.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[frob-group]
\begin{definition*}[Frobenius group]
  \glsadjdefn{frob_gp}{Frobenius}{group}
  \cloze{A \emph{Frobenius group} is a finite group $G$
  that has a transitive action on a set $X$ with
  $1 < |X| < |G|$ such that each $g \in G
  \setminus \{e\}$ fixes at most one element of
  $X$. It follows from \nameref{frob_1901} that
  Frobenius groups can't be simple. The subgroup
  $K$ is called the \emph{Frobenius kernel} and
  any of the subgroups
  \[ \Stab_G(x) \]
  for $x \in X$ are called \emph{Frobenius
  complements}.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $G = D_{2n}$. For $n$ odd acting on
      vertices of an $n$-gon in the usual way. The
      reflections fix precisely one vertex and the
      non-trivial solutions fix no vertices.
    \item \[ G = \left\{
      \begin{pmatrix}
        a & b \\
        0 & 1
      \end{pmatrix}
      \bigg| a, b \in \FF_p, a \neq 0
      \right\} \]
      acting on
      \[ X = \left\{
      \begin{pmatrix}
        x \\
        1
      \end{pmatrix}
      \bigg| x \in \FF_p
      \right\} \]
      by matrix multiplication.
  \end{enumerate}
\end{example*}

\begin{note*}
  No proof of \nameref{frob_1901} is known that
  does not use \gls{rep} theory!
\end{note*}

\begin{flashcard}[proof-of-frob-1901]
\begin{proof}[Proof of \nameref{frob_1901}]
  \cloze{Fix $x \in X$ and let $H = \Stab_G(x)$ so $|G| =
  |H| |X|$ by Orbit Stabiliser theorem. By
  hypothesis if $g \in G \setminus H$ then
  \[ \{e\} = \Stab_G(gx) \cap \Stab_G(x) .\]
  Thus
  \begin{enumerate}[(i)]
    \item $\left| \bigcup_{g \in G} gHg^{-1}
      \right| = \left| \bigcup_{x \in X}
      \Stab_G(x) \right| = (|H| - 1) |X| + 1$
    \item If $h_1, h_2 \in H$ then $[h_1]_H =
      [h_2]_H \iff [h_1]_G = [h_2]_G$.
    \item $C_G(h) = C_H(h)$ if $h \in H \setminus
      \{e\}$.
  \end{enumerate}
  By (i),
  \[ |K| = \left|\{e\} \cup \left(G \setminus
  \bigcup_{x \in X} \Stab_G(x) \right) \right| =
  |H| |X| - ((|H| - 1)|X| + 1) + 1 = |X| \]
  as claimed. We must show $K \normalsub G$.

  If $\chichar$ is any \gls{char_rep} of $H$, we
  can compute $\Ind_H^G \chichar$:
  \begin{align*}
    \Ind_H^G \chichar(g)
    &= \sum_{[h]_H \subset
    [g]_G} \frac{|C_G(x)|}{|C_H(h)|} \chichar(h)
    \\
    &= \begin{cases}
      \frac{|G|}{|H|} \chichar(e) & \text{if $g =
      e$} \\
        \chichar(h) & \text{if $[g]_G = [h]_G \neq
        \{e\}$ by (i) and (ii)} \\
        0 & \text{if $g \in K \setminus \{e\}$}
    \end{cases}
  \end{align*}
  Suppose $\Irr(H) = \{\chichar_1, \ldots,
  \chichar_r\}$ and let
  \[ \boxed{\theta_i = \Ind_H^G
  \chichar_i - \chichar_i(e) \indicator{G} -
  \chichar_i(e)\Ind_H^G \indicator{H} \in
  \charring(G)} \]
  (this is sort of the magic bit). So
  \[ \theta_i(g) = \begin{cases}
    \chichar_i(e) & g = e \\
    \chichar_i(h) & \text{$[g]_G = [h]_G$ for
    some $h \in H$} \\
    \chichar_i(e) & g \in K
  \end{cases} \]
  If $\theta_i$ were a \gls{char_rep} of a
  \gls{rep} of $G$ then the kernel of the
  \gls{rep} would contain $K$. Since $\theta_i \in
  \charring(G)$, $\theta_i = \sum n_i \psi_i$ for
  $n_i \in \ZZ$ and $\psi_i \in \Irr(G)$. Now we
  calculate:
  \begin{align*}
    \Gip\langle \theta_i, \theta_i \rangle_G
    &= \frac{1}{|G|} \sum_{g \in G}
    |\theta_i(g)|^2 \\
    &= \frac{1}{|G|} \left( \sum_{h \in H
    \setminus \{e\}} \frac{|G|}{|H|}
    |\chichar_i(h)|^2 + \sum_{k \in K}
    |\chichar_i(e)|^2 \right) \\
    &= \frac{|X|}{|G|} \sum_{h \in H}
    |\chichar_i(h)|^2 \\
    &= \langle \chichar_i, \chichar_i \rangle_H \\
    &= 1
  \end{align*}
  So $\sum n_j^2 = 1$ and $\theta_i = \pm \psi_j$
  for some $j$. But $g_i(e) = \chichar_i(e) > 0$
  so $\theta_i \in \Irr(G)$.

  To finish we write
  \[ \theta = \sum_{i = 1}^r \chichar_i(e)
  \theta_i \]
  and so $\theta(h) = \sum_{i = 1}^r \chichar_i(e)
  \chichar_i(h) = 0$ for $h \in H \setminus
  \{e\}$ by column orthogonality. Also $\theta(k)
  =\sum_{i = 1}^r \chichar_i(e)^2 = |H|$ by column
  orthogonality. Thus $K$ is the kernel of the
  \gls{rep} corresponding to $\theta$.}
\end{proof}
\end{flashcard}

In his thesis, John Thompson proved (among other
things), that $K$ must be nipotent or
equivalently the product of its Sylow-$p$-subgroups.