## Section14.1The Sylow Theorems

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 13. A group $G$ acts on itself by conjugation via the map $(g,x) \mapsto gxg^{-1}\text{.}$ 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)]\text{,} \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 $x_i\text{.}$

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

###### Proof.

We will use induction on the order of $G\text{.}$ If $|G|=p\text{,}$ then clearly $G$ itself is the required subgroup. We now assume that every group of order $k\text{,}$ where $p \leq k \lt n$ and $p$ divides $k\text{,}$ has an element of order $p\text{.}$ Assume that $|G|= n$ and $p \mid n$ and consider the class equation of $G\text{:}$

\begin{equation*} |G| = |Z(G)| + [G: C(x_1) ] + \cdots + [ G: C(x_k)]\text{.} \end{equation*}

We have two cases.

###### Case 1.

Suppose the order of one of the centralizer subgroups, $C(x_i)\text{,}$ is divisible by $p$ for some $i\text{,}$ $i = 1, \ldots, k\text{.}$ In this case, by our induction hypothesis, we are done. Since $C(x_i)$ is a proper subgroup of $G$ and $p$ divides $|C(x_i)|\text{,}$ $C(x_i)$ must contain an element of order $p\text{.}$ Hence, $G$ must contain an element of order $p\text{.}$

###### Case 2.

Suppose the order of no centralizer subgroup is divisible by $p\text{.}$ Then $p$ divides $[G:C(x_i)]\text{,}$ the order of each conjugacy class in the class equation; hence, $p$ must divide the center of $G\text{,}$ $Z(G)\text{.}$ Since $Z(G)$ is abelian, it must have a subgroup of order $p$ by the Fundamental Theorem of Finite Abelian Groups. Therefore, the center of $G$ contains an element of order $p\text{.}$

###### Example14.3.

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

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.

We induct on the order of $G$ once again. If $|G| = p\text{,}$ then we are done. Now suppose that the order of $G$ is $n$ with $n \gt p$ and that the theorem is true for all groups of order less than $n\text{,}$ where $p$ divides $n\text{.}$ We shall apply the class equation once again:

\begin{equation*} |G| = |Z(G)| + [G: C(x_1) ] + \cdots + [ G: C(x_k)]\text{.} \end{equation*}

First suppose that $p$ does not divide $[G:C(x_i)]$ for some $i\text{.}$ Then $p^r \mid |C(x_i)|\text{,}$ since $p^r$ divides $|G| = |C(x_i)| \cdot [G:C(x_i)]\text{.}$ Now we can apply the induction hypothesis to $C(x_i)\text{.}$

Hence, we may assume that $p$ divides $[G:C(x_i)]$ for all $i\text{.}$ Since $p$ divides $|G|\text{,}$ the class equation says that $p$ must divide $|Z(G)|\text{;}$ hence, by Cauchy's Theorem, $Z(G)$ has an element of order $p\text{,}$ say $g\text{.}$ Let $N$ be the group generated by $g\text{.}$ Clearly, $N$ is a normal subgroup of $Z(G)$ since $Z(G)$ is abelian; therefore, $N$ is normal in $G$ since every element in $Z(G)$ commutes with every element in $G\text{.}$ Now consider the factor group $G/N$ of order $|G|/p\text{.}$ By the induction hypothesis, $G/N$ contains a subgroup $H$ of order $p^{r- 1}\text{.}$ The inverse image of $H$ under the canonical homomorphism $\phi : G \rightarrow G/N$ is a subgroup of order $p^r$ in $G\text{.}$

A Sylow $p$-subgroup $P$ of a group $G$ is a maximal $p$-subgroup of $G\text{.}$ To prove the other two Sylow Theorems, we need to consider conjugate subgroups as opposed to conjugate elements in a group. For a group $G\text{,}$ let ${\mathcal S}$ be the collection of all subgroups of $G\text{.}$ For any subgroup $H\text{,}$ ${\mathcal S}$ 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 ${\mathcal S}\text{.}$

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 $G\text{.}$ Notice that $H$ is a normal subgroup of $N(H)\text{.}$ In fact, $N(H)$ is the largest subgroup of $G$ in which $H$ is normal.

###### Proof.

Certainly $x \in N(P)\text{,}$ and the cyclic subgroup, $\langle xP \rangle \subset N(P)/P\text{,}$ has as its order a power of $p\text{.}$ By the Correspondence Theorem there exists a subgroup $H$ of $N(P)$ containing $P$ such that $H/P = \langle xP \rangle\text{.}$ Since $|H| = |P| \cdot |\langle xP \rangle|\text{,}$ the order of $H$ must be a power of $p\text{.}$ However, $P$ is a Sylow $p$-subgroup contained in $H\text{.}$ Since the order of $P$ is the largest power of $p$ dividing $|G|\text{,}$ $H=P\text{.}$ Therefore, $H/P$ is the trivial subgroup and $xP = P\text{,}$ or $x \in P\text{.}$

###### Proof.

We define a bijection between the conjugacy classes of $K$ and the right cosets of $N(K) \cap H$ by $h^{-1}Kh \mapsto (N(K) \cap H)h\text{.}$ To show that this map is a bijection, let $h_1, h_2 \in H$ and suppose that $(N(K) \cap H)h_1 = (N(K) \cap H)h_2\text{.}$ Then $h_2 h_1^{-1} \in N(K)\text{.}$ Therefore, $K = h_2 h_1^{-1} K h_1 h_2^{-1}$ or $h_1^{-1} K h_1 = h_2^{-1} K h_2\text{,}$ and the map is an injection. It is easy to see that this map is surjective; hence, we have a one-to-one and onto map between the $H$-conjugates of $K$ and the right cosets of $N(K) \cap H$ in $H\text{.}$

###### Proof.

Let $P$ be a Sylow $p$-subgroup of $G$ and suppose that $|G|=p^r m$ with $|P|=p^r\text{.}$ Let

\begin{equation*} {\mathcal S} = \{ P = P_1, P_2, \ldots, P_k \} \end{equation*}

consist of the distinct conjugates of $P$ in $G\text{.}$ By Lemma 14.6, $k = [G: N(P)]\text{.}$ Notice that

\begin{equation*} |G|= p^r m = |N(P)| \cdot [G: N(P)]= |N(P)| \cdot k\text{.} \end{equation*}

Since $p^r$ divides $|N(P)|\text{,}$ $p$ cannot divide $k\text{.}$

Given any other Sylow $p$-subgroup $Q\text{,}$ we must show that $Q \in {\mathcal S}\text{.}$ Consider the $Q$-conjugacy classes of each $P_i\text{.}$ Clearly, these conjugacy classes partition ${\mathcal S}\text{.}$ The size of the partition containing $P_i$ is $[Q :N(P_i) \cap Q]$ by Lemma 14.6, and Lagrange's Theorem tells us that $|Q| = [Q :N(P_i) \cap Q] |N(P_i) \cap Q|\text{.}$ Thus, $[Q :N(P_i) \cap Q]$ must be a divisor of $|Q|= p^r\text{.}$ Hence, the number of conjugates in every equivalence class of the partition is a power of $p\text{.}$ However, since $p$ does not divide $k\text{,}$ one of these equivalence classes must contain only a single Sylow $p$-subgroup, say $P_j\text{.}$ In this case, $x^{-1} P_j x = P_j$ for all $x \in Q\text{.}$ By Lemma 14.5, $P_j = Q\text{.}$

###### Proof.

Let $P$ be a Sylow $p$-subgroup acting on the set of Sylow $p$-subgroups,

\begin{equation*} {\mathcal S} = \{ P = P_1, P_2, \ldots, P_k \}\text{,} \end{equation*}

by conjugation. From the proof of the Second Sylow Theorem, the only $P$-conjugate of $P$ is itself and the order of the other $P$-conjugacy classes is a power of $p\text{.}$ Each $P$-conjugacy class contributes a positive power of $p$ toward $|{\mathcal S}|$ except the equivalence class $\{ P \}\text{.}$ Since $|{\mathcal S}|$ is the sum of positive powers of $p$ and $1\text{,}$ $|{\mathcal S}| \equiv 1 \pmod{p}\text{.}$

Now suppose that $G$ acts on ${\mathcal S}$ by conjugation. Since all Sylow $p$-subgroups are conjugate, there can be only one orbit under this action. For $P \in {\mathcal S}\text{,}$

\begin{equation*} |{\mathcal S}| = |\text{orbit of }P| = [G : N(P)] \end{equation*}

by Lemma 14.6. But $[G : N(P)]$ is a divisor of $|G|\text{;}$ consequently, the number of Sylow $p$-subgroups of a finite group must divide the order of the group.

### Subsection14.1.1Historical Note

Peter Ludvig Mejdell Sylow was born in 1832 in Christiania, Norway (now Oslo). After attending Christiania University, Sylow taught high school. In 1862 he obtained a temporary appointment at Christiania University. Even though his appointment was relatively brief, he influenced students such as Sophus Lie (1842–1899). Sylow had a chance at a permanent chair in 1869, but failed to obtain the appointment. In 1872, he published a 10-page paper presenting the theorems that now bear his name. Later Lie and Sylow collaborated on a new edition of Abel's works. In 1898, a chair at Christiania University was finally created for Sylow through the efforts of his student and colleague Lie. Sylow died in 1918.