Problem Set 99: $p$-Fixed Point Congruence and Sylow's Theorems

Prove the $p$-Fixed Point Congruence Theorem.

Let $G$ be a finite group and let $p$ be a prime dividing $|G|$. Let $\sigma = (123\cdots p) \in S_p$ be a $p$-cycle in the $p$th symmetric group. Note that $C_p = \langle \sigma \rangle$ acts naturally on $p$-tuples $G^p$ by cyclically permuting coordinates. \[ \sigma (g_1,g_2,\ldots, g_p) := (g_2,g_3,\ldots, g_p, g_1). \]

1. Let $X \subseteq G^p$ be the set of all $p$-tuples whose coordinates multiply to $1$ in $G$, \[ X = \{ (g_1, g_2, \ldots, g_p) \in G^p : g_1g_2\cdots g_p = 1\}. \] Prove that $\sigma(X) = X$, hence $X$ is a $C_p$-set.
2. Prove that the number of elements in $X$ is $|X| = |G|^{p-1}$.
3. Prove that $\sigma$ must have at least 2 fixed points in $X$.

4. Prove that $G$ contains an element with order $p$.

   The theorem we just proved is due to Cauchy.

   Theorem 99.3 (Cauchy's Theorem)

   If $G$ is a finite group and $p$ is a prime dividing the order of $G$, then $G$ has an element of order $p$.

5. Let's prove a corollary. Let $p$ be a prime and suppose that $G$ is a non-trivial finite group such that every nontrivial element has $p$-power order. Prove that $G$ is a $p$-group.

If $G$ is a group, then the center of $G$, denoted $Z(G)$ is defined by \[ Z(G) = \{h \in G : ghg^{-1} = h \text{ for all $g \in G$}\}. \] The center is the subgroup of $G$ comprised of all elements which commute with every element of $G$. The following result may be proved using the $p$-fixed point congruence.

1. If $G$ is a $p$-group, prove that $Z(G)$ is a nontrivial subgroup. Hint: It all comes down to choosing the right action.
2. (Challenge) Let $p$ be a prime. Prove that any group of order $p^2$ is isomorphic to $C_{p^2}$ or $C_p \times C_p$.

Let $p$ be a prime. Let $\sigma = (12\cdots p) \in S_p$ be a $p$-cycle.

1. Let $0 < k < p$. Let $X_p := \{1,2,\ldots, p\}$ and let $\binom{X_p}{k}$ denote the set of all $k$ element subsets of $X_p$. Note that $C_p := \langle \sigma \rangle$ acts naturally on $\binom{X_p}{k}$. Use the $p$-fixed point congruence to prove that $\binom{p}{k}$ is a multiple of $p$. This can be done directly from the formula for $\binom{p}{k}$, but there is a richer proof using the $p$-fixed point congruence and I'd like you to find it.

2. Let $m$ be a natural number coprime to $p$. Prove that $\binom{mp^n}{p^n}$ is not a multiple of $p$ for any $n\geq 0$. Hint: Consider the group $G \subseteq S_{mp^n}$ generated by the disjoint $p^n$ cycles

   \[\begin{aligned} \sigma_1 &= (12\cdots p^n),\\ \sigma_2 &= ((p^n+1)(p^n+2)\cdots (2p^n)),\\ &\hspace{.55em}\vdots\\ \sigma_m &= (((m-1)p^n + 1)((m-1)p^n + 2)\cdots (mp^n)). \end{aligned}\]

   Since each of the $\sigma_i$ commute, each element $g \in G$ may be expressed as \[ g = \sigma_1^{a_1}\sigma_2^{a_2}\cdots \sigma_m^{a_m} \] for some integers $a_i$.

3. (Challenge) Suppose that $m, n$ are natural numbers with base-$p$ expansions

   \[\begin{aligned} m &= a_dp^d + a_{d-1}p^{d-1} + \ldots + a_1p + a_0\\ n &= b_dp^d + b_{d-1}p^{d-1} + \ldots + b_1p + b_0. \end{aligned}\]

   Note that this means each of the digits satisfy $ 0 \leq a_i, b_i < p$ for all $i$. Prove that \[ \binom{m}{n} \equiv \binom{a_d}{b_d}\binom{a_{d-1}}{b_{d-1}}\cdots \binom{a_1}{b_1}\binom{a_0}{b_0} \bmod p. \] This result is known as Lucas's Theorem; it is one of my favorites.

We will prove the following theorem.

1. Let $k \geq 0$ and define $\binom{G}{k}$ to be the set of all $k$-element subsets of $G$. Note that $G$ acts naturally on $\binom{G}{k}$ by left-multiplication. Prove that $\binom{G}{p^n}$ contains an element $Y$ such that $p$ does not divide the size of the orbit $GY$.
2. Let $Y$ be the subset from the previous part. Prove that the stabilizer $G_Y$ has order $p^n$.

Let $G$ be a finite group. Suppose $p$ is a prime such that $|G| = mp^n$ where $n\geq 1$ and $m$ is coprime to $p$.

1. Prove Sylow's second theorem.

   Theorem 99.5 (Second Sylow Theorem)

   If $H$ is a Sylow subgroup of $G$ and $K$ is any $p$-subgroup of $G$, then there exists a $g \in G$ such that $g K g^{-1} \subseteq H$. In particular, any two $p$-Sylow subgroups in $G$ are conjugate.

   Hint: Consider the natural action of $K$ on $G/H$.

2. If $H \subseteq G$ is a subgroup, the normalizer $N_G(H) \subseteq G$ of $H$ is the subgroup of all elements $g \in G$ such that $gHg^{-1} = H$. Note that $N_G(H)$ is the largest subgroup of $G$ for which $H$ is a normal subgroup of $N_G(H)$. Prove Sylow's third theorem.

   Theorem 99.6

   If $H \subseteq G$ is a $p$-Sylow subgroup and $m_p$ is the number of $p$-Sylow subgroups in $G$, then

   1. $m_p = [G : N_G(H)]$,
   2. $m_p \equiv 1 \bmod p$,
   3. $m_p$ divides $m$.

Let $p < q$ be primes and suppose that $q \not\equiv 1 \bmod p$. We will prove that any group $G$ of order $pq$ must be cyclic.

1. Prove that $G$ contains elements $g, h \in G$ with $\ord(g) = p$ and $\ord(h) = q$ such that $hgh^{-1} = g^k$ for some integer $k$.
2. Prove that $k^q \equiv 1 \bmod p$ and use this to argue that $gh = hg$.
3. Prove that $G = \langle gh\rangle$.
4. Classify all groups of order 15.

In this problem we practice using Sylow's Theorems.

1. Prove that any group of order 126 must contain a normal subgroup of order 7.
2. Suppose $G$ is a group with order 1000. Prove that $G$ contains a normal subgroup $N \tr G$ such that $1 < |N| < 1000$.