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)
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)
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)
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)
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 $G=N_G(P)\,H$.