Sylow Theorems

Let $G$ be a finite group. Recall the class formula:

$\displaystyle |G| = |Z(G)| + \sum_{|\mathcal O_x| > 1} [G : G_x].$

Problem 1. Let $p$ be a prime number such that $p \mid |G|$ . Use induction on $|G|$ to show that there exists a subgroup of $G$ with order $p^\alpha$ , where $p^\alpha \mid |G|$ and $p^{\alpha+1} \nmid |G|$ . We call such a subgroup a $p$ -Sylow subgroup of $G$ .

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Solution. The result is vacuously true for $|G| = 1$ . Suppose the result holds for any group $H$ with $|H| < |G|$ . Write

$|G| = p^\alpha \cdot m,\quad \gcd(p, m) = 1.$

In particular, $p^{\alpha+1} \nmid |G|$ . There are two cases: $p \mid |Z(G)|$ or $p \nmid |Z(G)|$ .

If $p \mid |Z(G)|$ , then $Z(G)$ contains an Abelian (and thus, normal) subgroup $H$ with order $p$ by Cauchy’s theorem, since $Z(G)$ is Abelian. Denote $\tilde G := G/H$ so that

$\displaystyle |\tilde G| = \frac{ |G| }{ |H| } = p^{\alpha - 1} \cdot m < |G|.$

By induction, $\tilde G$ contains a subgroup $K$ with order $p^{\alpha-1}$ . Now, the projection map $\pi : G \to \tilde G$ defined by $g \mapsto gH$ is a homomorphism, so that $\pi^{-1}(K) \triangleleft G$ . By definition,

$\pi^{-1}(K)/H = K.$

Therefore,

$\displaystyle \frac{ |\pi^{-1}(K)| }{ |H| }= |\pi^{-1}(K) / H| = |K|.$

Consequently, $|\pi^{-1}(K)| = |H| \cdot |K| = p \cdot p^{\alpha-1} = p^{\alpha}$ , as required.

Suppose $p \nmid |Z(G)|$ . By the class formula, write

$\displaystyle |G| = |Z(G)| + \sum_{i=1}^n [G : G_{x_i}]$

without loss of generality, where $|\mathcal O_{x_i}| > 1$ . If $p \mid [G : G_{x_i}]$ for each $i$ , then $p \mid |Z(G)|$ , a contradiction. Therefore, there exists some $i$ such that $p \nmid [G : G_{x_i}]$ . Note that $G_{x_i} \leq G$ and $[G : G_{x_i}] > 1$ , implies that

$\displaystyle |G_{x_i}| = \frac{ |G| }{ [G : G_{x_i}] } = p^\alpha \cdot \frac{ m }{ [G : G_{x_i}] } < |G|.$

By induction, $H$ contains a subgroup $K$ with order $p^{\alpha}$ . In particular, $K \leq H \leq G$ , as required.

Definition 1. Call a subgroup $H \leq G$ a $p$ -subgroup of $G$ if there exists an integer $n$ such that $|H| = p^n \mid |G|$ .

Problem 2. For any $p$ -Sylow subgroup $P$ , define

$\mathcal S := \{gPg^{-1} : g \in G\} = \{g_1 Pg_1^{-1}, \dots, g_n P g_n^{-1}\}.$

Show that $n \equiv_p 1$ .

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Solution. Define the collection

$\mathcal S := \{gPg^{-1} : g \in G\} = \{g_1 Pg_1^{-1}, \dots, g_n P g_n^{-1}\},$

since $G$ is finite. For any subgroup $Q$ , define the conjugation map $Q \times \mathcal S \to \mathcal S$ by

$(v, g_i P g_i^{-1}) \mapsto v (g_i P g_i^{-1}) v^{-1} = (vg_i)P(vg_i)^{-1}.$

Write $\mathcal S$ as a disjoint union of orbits under this action:

$\displaystyle \mathcal S := \bigsqcup_{j=1}^m \mathcal O_j, \quad \sum_{j=1}^m |\mathcal O_j| = n.$

Suppose for simplicity and without loss of generality that $g_j P g_j^{-1} \in \mathcal O_j$ for $1 \leq j \leq m$ . Then $|\mathcal O_j| = [Q : Q_{g_j P g_j^{-1}}]$ , where

$Q_H := \{ v \in Q : vHv^{-1} = H \}.$

By construction,

$Q_{g_j P g_j^{-1}} = G_{g_j P g_j^{-1}} \cap Q = g_j P g_j^{-1} \cap Q.$

Therefore, $|\mathcal O_j| = [Q : g_j P g_j^{-1} \cap Q]$ .

Now, we can show that $o(g_j P g_j^{-1}) = o(P)$ , so that each $g_j P g_j^{-1}$ is also a $p$ -Sylow subgroup. In particular, $o(g_1 P g_1^{-1}) = o(P)$ , so that $g_1 P g_1^{-1}$ is a $p$ -Sylow subgroup, and hence, a $p$ -subgroup. Replacing $Q$ with $g_1 P g_1^{-1}$ ,

$\displaystyle |\mathcal O_j| = [g_1 P g_1^{-1} : g_1 P g_1^{-1} \cap g_j P g_j^{-1}].$

In particular, $|\mathcal O_i| = 1$ , and

$g_1 P g_1^{-1} \neq g_j P g_j^{-1},\quad j \neq 1,$

so that $|\mathcal O_j| > 1$ . Since $P$ is a $p$ -Sylow subgroup, $g_j P g_j^{-1}$ is a $p$ -subgroup, so that there exists $\alpha_i$ such that $|g_j P g_j^{-1}| = p^{\alpha_i}$ . In particular, $p \mid p^{\alpha_j} \mid |\mathcal O_j|$ . Therefore,

$\displaystyle n = |\mathcal O_1| + \sum_{j=2}^m |\mathcal O_j| = 1 + kp.$

Therefore, $n \equiv_p 1$ .

Problem 3. Show that for any $p$ -subgroup $Q$ of $G$ , there exists a $p$ -Sylow subgroup $P$ such that $Q \leq P$ .

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Solution. Let $Q$ be any $p$ -subgroup of $G$ . By Theorem 1, there exists a $p$ -Sylow subgroup $P$ . Define $\mathcal S$ as per Problem 2. Suppose $Q \not\subseteq g_i P g_i^{-1}$ for any $i$ . Then $Q \cap g_i P g_i^{-1} \leq Q$ and $Q \cap g_i P g_i^{-1} \neq Q$ , so defining the orbits as in Problem 2, for any $1 \leq j \leq m$ ,

$|\mathcal O_j| = [Q : Q \cap g_j P g_j^{-1}] > 1.$

In particular, $p \mid |\mathcal O_j|$ for all $j$ , so that $n \equiv_p 0$ , a contradiction. Therefore, $Q \leq g_j P g_j^{-1}$ , and since the latter is a $p$ -Sylow subgroup, we are done.

Problem 4. Show that for any two $p$ -Sylow subgroups $H_1, H_2$ of $G$ , there exists $g \in G$ such that $H_2 = gH_1 g^{-1}$ . In this case, we say that $H_1,H_2$ are conjugate.

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Solution. Fix $p$ -Sylow subgroups $H_1, H_2$ . Write $P := H_1$ and define

$\mathcal S = \{g_1 P g_1^{-1}, \dots, u_n P u_n^{-1}\}$

as in Problem 2. Since $H_2$ is a $p$ -subgroup, there exists $j$ such that

$H_2 \leq g_j P g_j^{-1}.$

In particular, $p^\alpha = |H_2| \mid |g_j P g_j^{-1}|$ . If $|g_j Pg_j^{-1}| = p^{\alpha+1}$ , then $p^{\alpha + 1} \mid |G|$ , a contradiction. Therefore,