Sylow Theorems

Let $p$ be a prime. We recall that a finite group is called a $p$-group if its order is a power of $p$. The Sylow theorems are concerned with certain $p$-groups that can be found inside any finite group of order a multiple of $p$. They are extremely important facts in the theory of finite groups, and also are perfect examples of the principles in the last Section.

We have already defined the notion of a $p$-Sylow subgroup of a finite group $G$, but we recall that if it is a group of order $p^k$, where $p^k$ is the largest power of $p$ dividing $|G|$; it is assumed that $p| |G|$. We observe that a $p$-Sylow subgroup $P$ is necessarily a maximal $p$-subgroup of $G$. This means that if $Q$ is a $p$-subgroup of $G$ and $Q \supseteq P$ then $Q = P$. The reason is that the order of $Q$ must divide $|G|$, and $p^k$ is the largest power of $P$ dividing $G$.

Theorem 1.6.1: Let $p$ divide the order of the finite group $G$. Then $G$ has at least one $p$-Sylow subgroup.

Proof. (Click to Expand/Collapse)

First suppose that $p| |Z (G) |$. Then $Z (G)$ contains an element $x$ of order $p$, and $\left\langle x \right\rangle$ is a normal subgroup of order $p$. If $|G| = p^k m$ where $\gcd (m, p) = 1$, then $G' = G / \left\langle x \right\rangle$ has order $p^{k - 1} m$, and by induction on $|G|$, we may assume that $G'$ has a subgroup $P'$ of order $p^{k - 1}$. If $\pi : G \longrightarrow G'$ denotes the projection map, then $\pi^{- 1} (P')$ is a subgroup of $G$ of order $p^k$, that is, a $p$-Sylow, and we are done.

Next suppose that $p \nmid |Z (G) |$. Then by the class equation (1.5.2), $p \nmid [G : C_G (g_i)]$ for some noncentral $g_i$. Thus $C_G (g_i)$ is a proper subgroup of $G$

Exercise 1.6.1: Describe the Sylow subgroups of $S_4$.

Lemma 1.6.1: Let $G$ be a finite group, and let $H$, $P$ be $p$-subgroups of $G$, and that $H \subseteq N_G (P)$. Then $H P$ is a $p$-group. Moreover if $P$ is a $p$-Sylow subgroup, then $H \subseteq P$.

Proof. (Click to Expand/Collapse)

Let $N = N_G (P)$. Then $P \vartriangleleft N$ and $H \subseteq N$ so by Proposition 1.1.7, $H P$ is a group with $P$ as a normal subgroup, and $H P / P \cong H / (H \cap P)$. This is a $p$-group since $H$ is. Thus we have a short exact sequence \[ 1 \longrightarrow P \longrightarrow H P \longrightarrow H / (H \cap P) \longrightarrow 1 \] where both $P$ and $H / (H \cap P)$ are $p$-groups, and so $H P$ is a $p$-group.

If $P$ is a Sylow subgroup, then since $P$ is a maximal $p$-subgroup of $G$, $H P = P$. Since $H \subseteq H P$, we see that $H \subseteq P$.

A key point is to reinterpet this Lemma in terms of the action of $H$ on the Sylow subgroups by conjugation.

Proposition 1.6.1: Let $G$ be a finite group and $H$ a $p$-subgroup of $G$. Let $X$ be the set of $p$-Sylow subgroups of $G$. Then the orbit $\mathcal{O}$ of a $p$-Sylow subgroup $P$ of $G$ in $X$ has cardinality one (i.e $\mathcal{O}=\{P\}$) if and only $H \subseteq P$. Otherwise, the cardinality of $P$ is a power $p^r$ with $r > 1$.

Proof. (Click to Expand/Collapse)

This is a consequence of Lemma 1.6.1 together with the observation that the stabilizer of $P$ in the action of $H$ on the $p$-Sylows is $N_H (P) = N_G (P) \cap H$. By Theorem 1.5.1, the size of the orbit $\mathcal{O}$ is the index of this stabilizer in $H$. Since $H$ is a $p$-group, this cardinality is a power of $p$, and equals $1$ if and only if $H = N_H (P)$, that is, if and only if $H \subseteq N_G (P)$. By Lemma 1.6.1, this is true if and only if $H \subseteq P$.

Theorem 1.6.2: (Sylow Theorem). Let $G$ be a finite group and $p$ a prime dividing $|G|$. Then the number $n$ of $p$-Sylow subgroups of $G$ is a divisor of $|G|$ and is $\equiv 1$ modulo $p$. They are all conjugate to each other. If $H$ is any $p$-subgroup of $G$, then $H$ is contained in a $p$-Sylow subgroup.

Proof. (Click to Expand/Collapse)

Let $P$ be any $p$-Sylow subgroup. We consider the action of $P$ on the set $S$ of conjugates of $P$. Obviously a $p$-Sylow subgroup contains $P$ if and only if it equals $P$, so Proposition 1.6.1 implies that $\{P\}$ is one orbit for this action, and every other orbit has cardinality a multiple of $p$. Thus $|S| \equiv 1$ modulo $p$.

Now let us show that any $p$-subgroup $H$ of $G$ is contained in $Q$ for some $Q \in S$. Indeed, we consider the action of $H$ on $Q$ by conjugation. Since $H$ is a $p$-group, each orbit has order the index of the stabilizer, which is a power of $p$; there must be at least one orbit consisting of a single element, since $|S| \equiv 1$ modulo $p$, and every orbit that is not a single element has cardinality a multiple of $p$. If $\{Q\}$ is a single orbit, then by Proposition 1.6.1, $H \subseteq Q$.

We have proved that every $p$-subgroup is contained in a conjugate of $P$. In particular, each $p$-Sylow subgroup is contained in, and therefore equals, a conjugate of $P$. This proves that the $p$-Sylow subgroups are all conjugate to each other, and so $S$ is the set of all $p$-Sylows; thus $n \equiv 1$ modulo $p$. Moreover, since the group acts transitively, the number of $p$-Sylows is equal to the index of the stabilizer $N_G (P)$ of a typical $p$-Sylow $P$, hence is a divisor of $|G|$.

Exercise 1.6.2: Suppose that $H$ is a normal subgroup of $G$ and that $P$ is a $p$-Sylow subgroup of $G$. Prove that $P\cap H$ is a $p$-Sylow subgroup of $H$. Give an example to show that the normality of $H$ is necessary.

Exercise 1.6.3: Suppose that $H$ is a subgroup of $G$ and that $P$ is a $p$-Sylow subgroup of $H$. Show that $G=N_G(P)\,H$.