# A finite group of order $pq$ cannot be simple.

Let $p$ and $q$ be prime numbers. I wish to prove that a finite group $G$ of order $pq$ cannot be simple.

Proof. Case 1: $p\not= q$. Case 2: $p=q$.

Consider the first case where $p\not= q$. Without loss of generality, we can assume $p<q$. Then by the third Sylow theorem, $|\textrm{Syl}_p(G)|$ divides $q$. Since $q$ is prime and $q>p$ then $|\textrm{Syl}_p(G)|=1$. Let $P$ be the unique Sylow $p$-subgroup. We know that $P$ is not the trivial subgroup because $p<q$ and $|P|=p^r\geq p>1$. Then, by the second Sylow theorem, $P$ conjugates with itself so $gPg^{-1}=P$ for all $g\in G$. Thus, $P$ is normal and nontrivial, so $G$ is not simple.

I don't know how to approach the 2nd case where $p=q$. I know that $|G|=p^2$ and a group of prime squared order must be abelian. Would this help? Also, is this a good approach to the proof as a whole?

Thanks.

• All subgroups of an abelian group are normal. – Tim Mar 5 '15 at 23:16
• How can I know that there is a nontrivial proper subgroup? – Patrick Shambayati Mar 5 '15 at 23:17
• a group of order $p^2$ has an element of order $p$. Take the subgroup generated by that element. – Tim Mar 5 '15 at 23:18
• Does this follow from Sylow's first theorem? – Patrick Shambayati Mar 5 '15 at 23:34
• It follows from Cauchy's theorem (for groups). – David Wheeler Mar 5 '15 at 23:38

A group of order $p^2$ has a subgroup $H$ of order $p$ (if it is cyclic and $G=\langle a\rangle$, then $H$ can be $\langle a^p\rangle$, and if it is not cyclic, then take the subgroup generated by any element $\neq 1$) . Since the group is abelian this subgroup is normal.
• Why must there be a subgroup of order $p$? – Patrick Shambayati Mar 5 '15 at 23:36
• This has been explained carefully in the answer. The point is that every element of $G$ has order $1,p$ or $p^{2}.$ – Geoff Robinson Mar 5 '15 at 23:40