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.