We will use what we have learned about group actions to prove the Sylow Theorems. Recall for a moment what it means for $G$ to act on itself by conjugation and how conjugacy classes are distributed in the group according to the class equation, discussed in Chapter 14. A group $G$ acts on itself by conjugation via the map $(g,x) \mapsto gxg^{-1}$. Let $x_1, \ldots, x_k$ be representatives from each of the distinct conjugacy classes of $G$ that consist of more than one element. Then the class equation can be written as \begin{equation*}|G| = |Z(G)| + [G: C(x_1) ] + \cdots + [ G: C(x_k)],\end{equation*} where $Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}$ is the center of $G$ and $C(x_i) = \{ g \in G : g x_i = x_i g \}$ is the centralizer subgroup of $x_i$.

We begin our investigation of the Sylow Theorems by examining subgroups of order $p$, where $p$ is prime. A group $G$ is a $p$-group if every element in $G$ has as its order a power of $p$, where $p$ is a prime number. A subgroup of a group $G$ is a $p$-subgroup if it is a $p$-group.

##### Example15.3

Let us consider the group $A_5$. We know that $|A_5| = 60 = 2^2 \cdot 3 \cdot 5$. By Cauchy's Theorem, we are guaranteed that $A_5$ has subgroups of orders $2$, $3$ and $5$. The Sylow Theorems will give us even more information about the possible subgroups of $A_5$.

We are now ready to state and prove the first of the Sylow Theorems. The proof is very similar to the proof of Cauchy's Theorem.

##### Proof

A Sylow $p$-subgroup $P$ of a group $G$ is a maximal $p$-subgroup of $G$. To prove the other two Sylow Theorems, we need to consider conjugate subgroups as opposed to conjugate elements in a group. For a group $G$, let ${\mathcal S}$ be the collection of all subgroups of $G$. For any subgroup $H$, ${\mathcal S}$ is a $H$-set, where $H$ acts on ${\mathcal S}$ by conjugation. That is, we have an action \begin{equation*}H \times {\mathcal S} \rightarrow {\mathcal S}\end{equation*} defined by \begin{equation*}h \cdot K \mapsto hKh^{-1}\end{equation*} for $K$ in ${\mathcal S}$.

The set \begin{equation*}N(H) = \{ g \in G : g H g^{-1} = H\}\end{equation*} is a subgroup of $G$ called the the normalizer of $H$ in $G$. Notice that $H$ is a normal subgroup of $N(H)$. In fact, $N(H)$ is the largest subgroup of $G$ in which $H$ is normal.