% vim: tw=50
% 31/01/2023 11AM

\begin{enumerate}[(i)]
    \setcounter{enumi}{1}
    \item We prove a stronger result:
        \begin{flashcard}[sylow-ii-lemma]
        \begin{lemma}
            If $P \in \Syl_p(G)$ and $Q \le G$ is
            a $p$-subgroup then \cloze{$Q \le gPg^{-1}$
            for some $g \in G$.}
        \end{lemma}
        \end{flashcard}
        \begin{flashcard}[proof-of-sylow-ii-lemma]
            \prompt{Proof of Sylow(ii)? \\}
        \begin{proof}
            \cloze{
            Let $Q$ act on the left cosets $G / P$
            by left multiplication, ie
            \[ q \cdot gP=  qgP \]
            By the orbit-stabiliser theorem, each
            orbit has size dividing $|Q|$ so
            either 1 or a multiple of $p$. Since
            $|G / P| = m$ is coprime to $p$, there
            exists orbit of size 1, i.e. there
            exists $g \in G$ such that $qgP = gP$
            for all $q \in Q$.
            \begin{align*}
                \implies g^{-1}qg
                &\in P \quad \forall q \in Q \\
                \implies Q
                &\le gPg^{-1} \qedhere
            \end{align*}
        }
        \end{proof}
        \end{flashcard}
    \item
        \begin{flashcard}[proof-of-sylow-iii]
            \prompt{Proof of Sylow(iii)? \\}
        \cloze{
        Let $G$ act on $\Syl_p(G)$ by
        conjugation. Sylow (ii) implies action is
        transitive. Then the orbit-stabiliser
        theorem implies
        \[ n_p = |\Syl_p(G)| ~\big|~ |G| \]
        Now let $P \in \Syl_p(G)$. Then $P$ acts
        on $\Syl_p(G)$ by conjugation. The orbits
        have size dividing $|P| = p^a$, so either
        1 or a multiple of $p$. To show $n_p
        \equiv 1 \pmod{p}$ it suffices to show
        that $\{P\}$ is the unique orbit of size
        1.

        \myskip
        If $\{Q\}$ is an orbit of size 1, then $P$
        normalizes $Q$, i.e. $P \le N_G(Q)$. Now
        $P$ and $Q$ are Sylow $p$-subgroups of
        $N_G(Q)$, hence by (ii) are conjugate in
        $N_G(Q)$, hence equal since $Q \normalsub
        N_G(Q)$. Thus $\{P\}$ is the unique orbit
        of size 1.
        }
        \end{flashcard}
\end{enumerate}

\newpage
\section{Matrix Groups}

Let $F$ be a field (for example $\CC$ or $\ZZ /
p\ZZ$). Let
\begin{align*}
    \GL_n(F)
    &\defeq \text{$n \times n$ invertible matrices
    with entries in $F$.} \\
    \SL_n(F)
    &\defeq \ker(\GL_n(F)
    \stackrel{\det}{\longrightarrow} F^\times)
    \normalsub \GL_n(F)
\end{align*}
Let $Z \normalsub \GL_n(F)$ be the subgroup of
scalar matrices.

\begin{definition*}
    \begin{align*}
        \PGL_n(F)
        &= \frac{\GL_n(F)}{Z} \\
        \PSL_n(F)
        &= \frac{\SL_n(F)}{Z \cap \SL_n(F)}
        \cong \frac{Z\SL_n(F)}{Z}
        \le \PGL_n(F)
    \end{align*}
\end{definition*}

\begin{hiddenflashcard}[PGL-n-(F)]
    \begin{definition*}
        $\PGL_n(F) = \cloze{\frac{GL_n(F)}{Z}}$
    \end{definition*}
\end{hiddenflashcard}

\begin{hiddenflashcard}[PSL-n-(F)]
    \begin{definition*}
        $\PSL_n(F) = \cloze{\frac{\SL_n(F)}{Z \cap
        \SL_n(F)}} \cong \cloze{\frac{Z
        \SL_n(F)}{Z}} \le \PGL_n(F)$
    \end{definition*}
\end{hiddenflashcard}

\begin{example}
    $G = \GL_n(\ZZ / p\ZZ)$. A list of $n$ vectors
    in $(\ZZ / p\ZZ)^n$ are columns of some $A \in
    G$ if and only if they are linearly
    independent. Thus
    \begin{align*}
        |G|
        &= \ub{(p^n - 1)}_{\text{first column}} \cdot
        \ub{(p^n - p)}_{\text{second column}}
        \cdots (p^n - p^2)
        \cdots \ub{(p^n - p^{p - 1})}_{\text{last
        column}} \\
        &= p^{1 + 2 + \cdots + (n - 1)} (p^n -
        1)(p^{n - 1} - 1) \cdots (p - 1) \\
        &= p^{n \choose 2} \prod_{i = 1}^n (p^i -
        1)
    \end{align*}
    So Sylow $p$-subgroups have size $p^{n \choose
    2}$. Let
    \[ U = \left\{
    \begin{pmatrix}
        1 & * & \cdots & * \\
        0 & 1 & \cdots & * \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots 1
    \end{pmatrix}
    \right\} \le G \]
    set of upper triangular matrices with 1's on the
    diagonal. Then $U \in \Syl_p(G)$, since there
    are ${n \choose 2}$ entries above the diagonal
    to fill and each can take $p$ values. Just as
    $\PGL_2(\CC)$ acts on $\CC \cup \{\infty\}$
    via M\"obius maps, $\PGL_2(\ZZ / p\ZZ)$ acts
    on $\ZZ / p\ZZ \cup \{\infty\}$. Indeed
    $\GL_2(\ZZ / p\ZZ$ acts as
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    : z \mapsto \frac{az + b}{cz + d} \]
    and since scalars act trivially, we obtain an
    action of $\PGL_2(\ZZ / p\ZZ)$.
\end{example}

\begin{hiddenflashcard}[size-of-GLn]
\[ |\GL_n(\ZZ/p\ZZ) = \cloze{(p^n - 1)
(p^n - p) \cdots (p^n - p^{n - 1})} \]
\end{hiddenflashcard}

\begin{flashcard}[PGL-permutation-representation]
\begin{lemma}
    The permutation representation $\PGL_2(\ZZ /
    p\ZZ) \cloze{\to S_{p + 1}}$ \cloze{is injective (in fact an
    isomorphism if $p = 2$ or $p = 3$).}
\end{lemma}

\begin{proof}
    \cloze{
    Suppose $\frac{az + b}{cz + d} = z$ for all $z
    \in \ZZ / p\ZZ \cup \{\infty\}$. Setting $z =
    0$ gives $b = 0$, $z = \infty$ gives $c = 0$,
    $z = 1$ gives $a = d$, so
    \[
    \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}
    \]
    is a scalar matrix, hence trivial in
    $\PGL_2(\ZZ / p\ZZ)$.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[size-of-PSL-2-(ZZp)]
\begin{lemma}
    If $p$ is an odd prime then
    \[ |\PSL_2(\ZZ / p\ZZ)| = \cloze{\frac{p(p - 1)(p +
    1)}{2}} \]
\end{lemma}
\end{flashcard}

\begin{proof}
    By Example 5.1
    \[ |\GL_2(\ZZ / p\ZZ)| = p(p - 1)(p^2 - 1) \]
    The group homomorphism
    \[ \GL_2(\ZZ / p\ZZ)
    \stackrel{\det}{\longrightarrow} (\ZZ /
    p\ZZ)^\times \]
    is surjective:
    \[
    \begin{pmatrix}
        a & 0 \\
        0 & 1
    \end{pmatrix}
    \mapsto a \]
    therefore $|\SL_2(\ZZ / p\ZZ) =
    \frac{\GL_2(\ZZ / p\ZZ)}{p - 1} = p(p - 1)(p +
    1)$. If
    \[
    \begin{pmatrix}
        \lambda & 0 \\
        0 & \lambda
    \end{pmatrix}
    \in \SL_2(\ZZ / p\ZZ) \]
    then $\lambda^2 \equiv 1 \pmod{p}$
    \begin{align*}
        \implies p
        &\mid (\lambda - 1)(\lambda + 1) \\
        \implies \lambda
        &\equiv \pm 1 \pmod{p}
    \end{align*}
    Thus $Z \cap \SL_2(\ZZ / p\ZZ) = \{\pm I\}$
    (distinct since $p > 2$). Thus
    \begin{align*}
        |\PSL_2(\ZZ / p\ZZ)|
        &= \half |\SL_2(\ZZ / p\ZZ)| \\
        &= \frac{p(p - 1)(p + 1)}{2}
        \qedhere
    \end{align*}
\end{proof}