## Section15.1The 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 14. 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.

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 $$p\text{.}$$

###### Example15.3.

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.

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.

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 $$x \in P\text{.}$$

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 $$H\text{.}$$

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 15.6, $$k = [G: N(P)]\text{.}$$ 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 15.6, 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 15.5, $$P_j = Q\text{.}$$

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 15.6. 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.

### SubsectionHistorical 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.