Group of order pq is not simple

Is the following correct way of showing that there is no simple group of order $pq$ where $p$ and $q$ are distinct primes?

If $|G|=n=pq$ then the only two Sylow subgroups are of order $p$ and $q$.

From Sylow's third theorem we know that $n_p | q$ which means that $n_p=1$ or $n_p=q$.

If $n_p=1$ then we are done (by a corollary of Sylow's theorem)

If $n_p=q$ then we have accounted for $q(p-1)=pq-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.

Is that correct?

Solution 1:

This CW answer intends to remove the question from the unanswered queue.

As already noted in the comments, your proof is correct.

Solution 2:

If |G|=pq where p and q are primes then without loss of generality assume p < q. By Sylow's theorems there exists a Sylow q subgroup. The number of these is congruent to 1 mod q but also divide p by Sylow's theorems. Now since q>p we can say there is only one sylow q subgroup. Using the fact that conjugate elements have the same order, the unique sylow q subgroup is normal.