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\newpage
\section{$p$-groups and $p$-subgroups}

\begin{definition*}
    Let $p$ be a prime. A finite group $G$ is a
    $p$-group if $|G| = p^n$, $n \ge 1$.
\end{definition*}

\begin{theorem}
    If $G$ is a $p$-group, then $Z(G) \neq 1$.
\end{theorem}

\begin{proof}
    For $g \in G$, we have $|\ccl_G(g)||C_G(g)| =
    |G| = p^n$, so each conjugacy class has size a
    power of $p$. Since $G$ is a union of
    conjugacy classes:
    \[ |G| = \#(\text{conjugacy classes of size
    $1$}) \pmod{p} \]
    Note that
    \begin{align*}
        g \in Z(G)
        &\iff gxg^{-1} = x ~\forall x \in G \\
        &\iff x^{-1}gx = g ~\forall x \in G \\
        &\iff \ccl_G(g) = \{g\}
    \end{align*}
    So $|Z(G)| = \#(\text{conjugacy classes of
    size 1})$.
    So $0 \equiv |Z(G)| \pmod{p}$. We know $|Z(G)| \ge
    1$ since $e \in Z(G)$, so therefore $|Z(G)|
    \ge p > 1$.
\end{proof}

\begin{corollary}
    The only simple $p$-group is $C_p$.
\end{corollary}

\begin{proof}
    Let $G$ be a simple $p$-group. Since $Z(G)
    \normalsub G$ we have $Z(G) = 1$ or $G$. But
    by the previous theorem, $Z(G) \neq 1$, so
    $Z(G) = G$, so $G$ is abelian. Conclude by
    Lemma 1.3.
\end{proof}

\begin{flashcard}[p-group-has-subgroups-of-every-order]
\begin{corollary*}
    Let $G$ be a $p$-group of order $p^n$. Then
    $G$ has \cloze{a subgroup of order $p^n$ for all $0
    \le r \le n$.
}
\end{corollary*}

\begin{proof}
    \cloze{
    By Lemma 1.4, $G$ has a composition series
    \[ 1 = G_0 \normalsub G_1 \normalsub \cdots
    \normalsub G_{m - 1} \normalsub G_m = G ,\]
    with each $G_i / G_{i - 1}$ being simple, and
    also since $G$ is a $p$-group, $G_i / G_{i -
    1}$ is a $p$-group, so $G_i / G_{i = 1} \cong
    C_p$ by Corollary 4.2.

    \myskip
    Thus $|G_i| = p^i$ for $0 \le i \le m$ and $m
    = n$.
}
\end{proof}
\end{flashcard}

\begin{lemma}
    For $G$ a group, if $G / Z(G)$ is cyclic, then
    $G$ is abelian (and so $G / Z(G)$ is trivial).
\end{lemma}

\begin{proof}
    Let $gZ(G)$ be a generator for $G / Z(G)$.
    Then each coset is of the form $g^r Z(G)$ for
    some $r \in \ZZ$. Thus $G = \{g^r z \colon r
    \in \ZZ, z \in G(Z)\}$. Then
    \begin{align*}
        (g^{r_1} z_1) \cdot (g^{r_2} z_2)
        &= g^{r_1 + r_2} z_1 z_2 \\
        &= g^{r_1 + r_2} z_2 z_1 \\
        &= (g^{r_2} z_2) \cdot (g^{r_1} z_1)
    \end{align*}
    So $G$ is abelian.
\end{proof}

\begin{corollary}
    If $|G| = p^2$, then $G$ is abelian.
\end{corollary}

\begin{proof}
    We consider the 3 possible cases for $|Z(G)|$
    ($|Z(G)| ~\big|~ p^2$ by Lagrange's theorem)
    \begin{itemize}
        \item If $|Z(G)| = 1$, then this
            contradicts Theorem 4.1.
        \item If $|Z(G)| = p$, then $|G / Z(G)| =
            p$. Apply Lemma 4.1, contradiction.
        \item $|Z(G)| = p^2$, then $Z(G) = G$ so
            $G$ is abelian. \qedhere
    \end{itemize}
\end{proof}

\myskip
See example sheet for case $|G| = p^3$.

\subsection{Sylow Theorems}

\begin{flashcard}[sylow-theorems]
\begin{theorem*}[Sylow]
Let $G$ be a finite group of order $p^a m$ where
$p$ is a prime with $p \nmid m$. Then
\begin{enumerate}[(i)]
    \item \cloze{The set $\Syl_p(G) = \{P \le G \colon
        |P| = p^a\}$ of Sylow $p$-subgroups is
        non-empty.}
    \item \cloze{All elements of $\Syl_p(G)$ are
        conjugate.}
    \item \cloze{$n_p \defeq |\Syl_p(G)|$ satisfies $n_p
        \equiv 1 \pmod{p}$ and $n_p ~\big|~ |G|$
        (and hence $n_p \mid m$).}
\end{enumerate}
\end{theorem*}
\end{flashcard}

\begin{corollary}
    If $n_p = 1$, then the unique Sylow
    $p$-subgroup is normal.
\end{corollary}

\begin{proof}
    Let $g \in G$ and $P \in \Syl_p(G)$. Then
    $gPg^{-1} \in \Syl_p(G)$ and so $gPg^{-1} =
    P$. Thus $p \normalsub G$.
\end{proof}

\begin{example*}
    Let $|G| = 1000 = 2^3 \times 5^3$. Then $n_5
    \equiv 1 \pmod{5}$ and $n_5 \mid 8$, so $n_5 =
    1$. Thus the unique Sylow $5$-subgroup is
    normal, and hence $G$ is not simple.
\end{example*}

\begin{example*}
    $|G| = 132 = 2^3 \times 3 \times 11$. $n_{11}
    \equiv 1 \pmod{11}$ and $n_{11} \mid 12$, so
    $n_{11} = 1$ or $n_{11} = 12$. Suppose $G$ is
    simple. Then $n_{11} \neq 1$ (otherwise the
    Sylow 11 subgroup is normal) and hence $n_{11}
    = 12$. Now $n_3 \equiv 1 \pmod{3}$ and $n_3
    \mid 44$. So $n_3 = 4, 22$ ($n_3 \neq 1$
    if $G$ is simple).

    \myskip
    Suppose $n_3 = 4$. Then letting $G$ act on
    $\Syl_3(G)$ by conjugation gives a group
    homomorphism $\phi \colon G \to S_4$. Since
    $G$ is simple, we must have $\Ker(\phi) =1$ or
    $\Ker(\phi) = G$. But $\Ker(\phi) = G$
    contradicts Sylow (ii). So $\Ker(\phi) = G$,
    so $G \injto S_4$. But this is not possible
    since $|G| > |S_4|$.

    \myskip
    Thus $n_3 = 22$ and $n_{11} = 12$. So $G$ has
    $22 \times (3 - 1) = 44$ elements of order 3
    and $12 \times (11 - 1) = 120$ elements of
    order 11. But $44 + 120 > 132 = |G|$.

    \myskip
    Hence there does not exist a simple group of
    order $132$.
\end{example*}

\subsubsection*{Proof of Sylow Theorems}

Let $|G| = p^a m$, $p$ prime, $p \nmid m$.
\begin{enumerate}[(i)]
    \item
        \begin{flashcard}[proof-of-sylow-i]
        \prompt{Proof of Sylow(i)? \\}
        \cloze{
        Let $\Omega$ be the set of all subsets
        of $G$ of size $p^a$.
        \[ |\Omega| = {p^a m \choose p^a} =
        \frac{p^a m}{p^a} \cdot \frac{p^a m -
        1}{p^a - 1} \cdots \frac{p^a m - p^a +
        1}{1} \]
        For $0 \le k < p^a$, the numbers $p^a m -
        k$ and $p^a - k$ are divisible by the same
        power of $p$. Therefore $|\Omega|$ is
        coprime to $p$ (\dag).

        \myskip
        Let $G$ act on $\Omega$ by left
        multiplication, i.e. for $g \in G$ and $X
        \in \Omega$
        \[ g * X = \{gx \colon x \in X\} \in
        \Omega \]
        For any $X \in \Omega$ we have $|G_X|
        |\orb_G(X)| = |G| = p^a m$. By (\dag) there
        exists $X$ such that $|\orb_G(X)|$ is
        coprime to $p$. Thus $p^a ~\big|~ |G_X|$
        (1).
        On the other hand, if $g \in G$ and $x \in
        X$, then $g \in (gx^{-1}) * X$ and hence
        \[ G = \bigcup_{g \in G} g * X =
        \bigcup_{Y \in \orb_G(X)} Y \]
        \[ \implies |G| \le |\orb_G(X)| |X| \]
        \[ \implies |G_X| =
        \frac{|G|}{|\orb_G(X)|} \le |X| = p^a
        \tag{2} \]
        (1) and (2) implies
        \[ |G_X| = p^a \]
        i.e. $G_X \in \Syl_p(G)$.
        }
        \end{flashcard}
\end{enumerate}