Existence of Sylow subgroups; the first Sylow Theorem

Definition 10.1.1   Let $G$ be a finite group with $\vert G\vert = n$ and let $p$ be a prime such that $p^a\vert n$ but no higher power of $p$ divides $n$. A subgroup of $G$ of order $p^a$ is called a $p$-Sylow subgroup.

It is not at all obvious that a $p$-Sylow subgroup exists. It is our main concern in this section to show that for each $p\vert n$ that a $p$-Sylow subgroup exists. Note that $P$ is a $p$-Sylow subgroup of $G$ if and only if $G = p^rn$ where $p\not\vert n$ and $\vert P\vert = p^r$.

We first consider and prove a very special case of the end result we wish to obtain.

Theorem 10.1.2   Let $G$ be a finite abelian group and let $p$ be a prime such that $p\vert\, \vert\, \vert G\vert$. Then $G$ contains at least one element of order $p$.

Proof: The proof will proceed by induction on $\vert G\vert$. If $\vert G\vert = p$, a prime, then the theorem is clearly true. Thus suppose $\vert G\vert = n$, where $n$ is composite and also suppose that the theorem has been proven for all groups whose order $< n$. Suppose $p$ is a prime such that $p\vert n$. We need to show that $G$ has an element of order $p$. We claim $G$ contains a subgroup not equal to $\{e\}$ or to $G$ itself. This is clear if $G$ is cyclic (see Theorem 5.2.1). If $G$ is not cyclic, let $a \in G$, $a \not= e$, the identity element of $G$. Then $\langle a\rangle$ is a proper subgroup of $G$. (Why?) Let $H$ be a proper subgroup of $G$ of maximal order. If $p\vert\,\vert H\vert$, then since $\vert H\vert < \vert G\vert$, we have by the induction hypothesis that there exists an $h\in H$ such that $o(h) = p$, but clearly $h \in G$, also. This proves our claim when $p\vert\,\vert H\vert$. If, however, $p\not\vert \vert H\vert$, then since $H$ is a proper subgroup of $G$, there exists an element $a \in G - H$. Let $K$ be the cyclic subgroup of $G$ generated by $a$, i.e., $K = \langle a\rangle$. Now the product $HK$ is a subgroup of $G$ (by Theorem 4.2.1) since $G$ is abelian. Also, $H \subset HK$ properly, i.e., $H \not= HK$, because $a \in HK$ but $a \in H$. However, $H$ was a maximal proper subgroup. Thus it must be that $HK = G$. Then, by Theorem 4.3.6,

\vert G\vert=\frac{\vert H\vert\vert K\vert}{\vert H\cap K\vert}=\frac{\vert H\vert\cdot o(a)}{d},

where $d = \vert H \cap \langle a\rangle$. Thus

d\vert G\vert = \vert H\vert\cdot o(a),

and since $p\vert\, \vert G\vert$, we must have that $p\vert\, \vert H\vert o(a)$. However, we have assumed $p \not\vert \vert H$, thus by the Corollary 1.2.10 we have $p\vert o(a)$. Let $o(a) = m$. Then $m = pk$, where $k\in \mathbb{N}$, and consider the element $a^k$. By Theorem 5.2.2,



Thus if $G$ is an abelian group and if $p\vert n$, then $G$ contains a subgroup of order $p$; viz., the cyclic subgroup of order $p$ generated by an element $a \in G$ of order $p$ whose existence is guaranteed by Theorem 10.1.2.

We now proceed to the main result of this section, i.e., the first Sylow Theorem.

Theorem 10.1.3 (Sylow I)   Let $G$ be a finite group and let $p \in G$, then $G$ contains a $p$-Sylow subgroup (i.e., a $p$-Sylow subgroup exists).

Proof: As in the preceding theorem, the proof will be given by induction on $\vert G\vert$. The theorem is clearly true if $\vert G\vert = 2$. Now let $\vert G\vert = n = p^an'$, where $p\not\vert n'$. By hypothesis, $a>0$. We next decompose $G$ into conjugacy classes according to the discussion at the beginning of Section 4.1 and use the class equation, i.e., equation (4.8), to obtain

G = Z(G) \cup Cl(a_1) \cup Cl(a_2)\cup ... \cup Cl(a_t)
\ \ \ \ ({\rm disjoint})
\end{displaymath} (10.1)

where $Cl(a_i)$ designates a conjugacy class. Since the union is disjoint, we can write

n = \vert Z(G)\vert + k_1 + k_2 + ... + k_t,

where $k_j = \vert Cl(a_j)\vert$. By Theorem 4.3.4, $k_j = [G : CG(a_j)] = n/n_j$, where $n_j = CG(a_j)$. We return now to equation (10.1) and recall that the conjugacy classes, $Cl(a_j)$, listed (if there are any) are nontrivial, i.e., each $k_j > 1$. Let us suppose that some $k_j$ is such that $p \not\vert k_j$, i.e., $gcd (k_j, p) = 1$. Since $n_jk_j = n$, we must have $n_j < n$. Moreover, $p^a\vert n_j$ since $p \not\vert k_j$. It follows by the induction hypothesis that the subgroup $CG(a_j)$ contains a $p$-Sylow subgroup and that therefore $G$ contains a $p$-Sylow subgroup. Thus, in this case, the theorem has been established.

We may, thus, assume that for each $j$, $j = 1, 2, ..., t$, $p \vert k_j$. Thus

p^an' = \vert Z(G)\vert + pr,

whence $p\vert\, \vert Z(G)\vert$. Since $Z(G)$ is an abelian group, we have by the preceding theorem that $Z(G)$, and therefore, $G$ has an element, $a$, of order $p$. Now $\langle a\rangle \lhd G$ since $a\in Z(G)$ and $\vert\langle a\rangle \vert = p$. Hence $\vert G/\langle a\rangle \vert = p^{a- 1}n'$, and so by the induction hypothesis $G/\langle a\rangle$ must contain a $p$-Sylow subgroup of order $p^{a- 1}$. This $p$-Sylow subgroup must be of the form $P/\langle a\rangle$, where $P \leq G$ which contains $\langle a\rangle$ by the Corollary 8.3.3. Now

\vert P\vert=\vert P \langle a\rangle \vert \cdot \vert\langle a\rangle \vert
p^{a- 1}\cdot p=p^a,

and so $P$ is a $p$-Sylow subgroup of $G$. $\Box$

On the basis of this theorem, we can now strengthen the result obtained in Theorem 10.1.2.

Theorem 10.1.4 (Cauchy)   If $G$ is a finite group and if $p$ is a prime such that $p\vert\, \vert G\vert$, then $G$ contains at least one element of order $p$.

Proof: Let $P$ be a $p$-Sylow subgroup of $G$, and let $P = p^a$. If $e \not= a \in P$, then $o(a)\vert\, \vert P\vert$ implies $o(a) = p^b$, where $0 <b\leq a$. But then the cyclic group, $\langle a\rangle$, must have a (unique) subgroup of order $p$, say $\langle a^t\rangle$, by Theorem 5.2.1. Thus $a^t \in G$ and $o(at) = p$. $\Box$


David Joyner 2007-08-06