Abstract Algebra: Theory and Applications

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 $p\le k<n$ 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 $G\text{:}$

We have two cases.

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 $n>p$ 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:

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 $\varphi :G\to G/N$ is a subgroup of order $p^r$ in $G\text{.}$

Certainly $x\in N(P)\text{,}$ and the cyclic subgroup, $⟨xP⟩\subset N(P)/P\text{,}$ 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 $H/P=⟨xP⟩\text{.}$ Since $|H|=|P|\cdot |⟨xP⟩|\text{,}$ 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

consist of the distinct conjugates of $P$ in $G\text{.}$ By Lemma 15.6, $k=[G:N(P)]\text{.}$ Notice that

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,

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 $1\text{,}$ $|\mathcal{S}|\equiv 1(\mathrm{mod}p)\text{.}$

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 $P\in \mathcal{S}\text{,}$

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.