# Every group has a subgroup of prime order?

Is there a quick proof that given any finite group $G$ with $|G| = n$, it has a subgroup of prime order $p \geq 2$?

I've managed to prove the statement by writing down the unique prime factorization of $n$, picking a prime $p$ in the factorization of $n$, applying the Sylow Theorem stating that $G$ has a subgroup which is a $p$-group and then arguing that every $p$-group has a subgroup of order $p$. However, this feels unnecessarily complicated, and I was wondering whether a simpler proof exists.

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Also note that one normally uses this to prove the Sylow theorems, so using Sylow to prove this really isn't the right thing to do. – Your Ad Here Oct 31 '12 at 19:59
The question as stated is very easy, assuming $n>1$. Just choose any non-identity element and raise it to a suitable power to get an element of prime order. But perhaps you are really looking for a subgroup of order $p$ for an arbitrary prime $p$ dividing $|G|$? – Derek Holt Oct 31 '12 at 20:03
@DerekHolt: It looks like the question is as stated. – P.. Nov 1 '12 at 22:26

Let $a \neq 1$ be an element of $G$. Let $H$ be the subgroup generated by $a$. Let $p$ be a prime number which divides $|H|$. Then $H$ contains a unique subgroup of order $p$.

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A proof for the abelian case:

Assume that $p||G|$.

It is enough to show that there exist $g \in G$ s.t. $p|o(g)$.We will use induction on $|G|$.

(*) If ${1}< N < G$ with $p||N|$ or $p||G/N|$ then by the inductive hypothesis we are done. For the latter, exist $g \in G$ s.t. $g^p=n \in N \Rightarrow p|o(g)$.

If there exist ${1}<N<G$ use Lagrange and (*). If not then $|G|=p$ and $G$ is cyclic. In either case we can find $g \in G$ with $p|o(g)$.

A proof for the general case due to James H. McKay:

Assume that $|G|=n$ and $p|n$ prime.

Define $A=\{(a_1,a_2,\ldots,a_p):a_1a_2\cdots a_p=1\}$ and $H=\{\sigma^i:i=1,2,\ldots,p\}$ where $\sigma=(1 \ 2 \ 3 \ \ldots \ p) \in S_p$. Then $|A|=n^{p-1}$ (If you pick $a_1, a_2, \ldots, a_{p-1}$ arbitrary then $a_n$ is fixed) and $|H|=p$. Define the equivalence relation in $A$, $$(a_1,a_2,\ldots,a_p)\sim (b_1,b_2,\ldots,b_p) \iff (a_1,a_2,\ldots,a_p)=(b_{h(1)},b_{h(2)},\ldots,b_{h(p)}) \text{ for some } h \in H.$$ Observe that the equivalence classes either contain $1$ element (being $(x,x,\ldots ,x)$) or $p$ elements. If we have $r$ equivalent classes with $1$ element ($r\geq 1$ since $(1,1,\ldots , 1)$ is one) and $q$ with $p$ elements then $$r+p\cdot q=n^{p-1}.$$ So $p|r$ and $r>1$. Therefore exist $x\neq 1$ s.t. $x^p=1.$

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But not all subgroups are normal! Your proof fails on nonabelian simple groups. – Derek Holt Nov 1 '12 at 15:23
You are right. The proof is only for abelian groups. I will add this. Thanks. – P.. Nov 1 '12 at 19:48

Let $|G|=mp$, Let us define a set $X=\{A \subset G | |A|=p\}$. Note that $|X|=\binom {np}{p}$. The largest power $k$ of $p$ that divides $|X|$ is same as the largest power of $p$ that divides $n$.

We have a $G$ action on $X$ as $g \times A= \{gx | x \in A\}$. We write the orbit decomposition for this action $|X| = \sum |G|/|G_A|$. Now note that $p^k | |X|$ so depending on $k$ we have the following two cases :

Case 1: if $k=0$ then $p$ does not divide $|X|$ that means there is atleast one $A \in X$ such that $p | G_A$. Now if $G_A= G$ then it $A$ is a group of order $p$ ! (Homework) else we have a smaller group of order divisible by $p$. So by induction we have a subgroup of order $p$.

Case 2: else $p^k | |X|$ so there must be atleast one $G_A$ for some $A \in X$ such that $p || G_A|$ since $p ^{k+1} | |G|$. Now again we have a smaller subgroup with order divisible by $p$.

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