# Groups of Order $pq$ and Cayley's Theorem

It's been a while since my last course on abstract algebra and I'm going through all of the exercises in Herstein's Topics in Algebra to refresh myself. This one comes from the section on Cayley's theorem. It's one of the harder starred problems:

"Let $o(G)$ be $pq$, $p>q$ are primes, prove $G$ has a subgroup of order $p$ and a subgroup of order $q$."

Now I know this can easily be proved with Sylow's theorem or Cauchy's theorem, but those topics aren't introduced until later on in this book.

If $G$ has an element $g$ of order $pq$, then $G$ must be cyclic, in which case $\langle g^q\rangle$ and $\langle g^p\rangle$ are subgroups of order $p$ and $q$, respectively.

We can then consider the case where $G$ is not cyclic. Since the order of $G$ is clearly not prime, $G$ must have a nontrivial subgroup, $H$. By Lagrange its order must be of order $p$ or $q$. I think the idea is to prove that the existence of one such subgroup implies the existence of the other.

If the order of $H$ is $p$, then its index in $G$ is $q$. One of the corollaries to Cayley's theorem says that since $o(G) \nmid [G:H]!$, $H$ must contain a nontrivial normal subgroup of $G$. This condition is met since $pq \nmid q!$ because $p>q$. Since $H$ itself has prime order, the only nontrivial subgroup of $G$ it contains is itself, so $H$ must be normal in $G$.

That's what I have so far. I would then try to use this to show the existence of a subgroup of order $q$. Then I would try to do the reverse and show that the existence of a subgroup of order $q$ implies the existence of another subgroup of order $p$. Is there a better way (that uses Cayley's theorem and not Sylow/Cauchy)?

Thank you!

• The proof of Cauchy's Theorem is very elementary. It is really natural to use it here. I also have seen questions about group so f order $pq$, where Lagrange's Theorem "is not allowed" to be used, or properties of a product of primes $pq$ are not allowed to be used. I think this is not reasonable. – Dietrich Burde Apr 15 '17 at 20:18

## 1 Answer

Suppose that all elements different from $e$ have order $p$.

Then, the union of all subgroups of order $p$ is the whole group. Moreover, any two such subgroups are either equal or have trivial intersection. Therefore, if $n$ is the number of subgroups of order $p$, then $n(p-1)+1=pq$ and so $$n=\frac{pq-1}{p-1}=q +\frac{q-1}{p-1}$$ But this is never an integer if $p>q \ge 2$. Therefore, there is at least one element of order $q$.

We still need to prove that there is at least one element of order $p$. Unfortunately, the argument cannot be repeated swapping $p$ and $q$.