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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?

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Looks fine to me. You can perhaps shorten the proof by arguing that $\,G\,$ must have a unique Sylow subgroup of order the highest prime among $\,p\,,\,q\,$... –  DonAntonio Oct 28 '12 at 17:14
    
In fact, these groups are not too difficult to fully classify. See, e.g., groupprops.subwiki.org/wiki/… –  Benjamin Dickman Oct 28 '12 at 17:30
    
You could also argue as follows: Since $|G|=pq$, Cauchy's Theorem implies that there exists an element $x$ of order $p$ and an element $y$ of order $q$. Since $p,q$ are prime, $\langle x \rangle $ and $\langle y \rangle $ are cyclic subgroups of order $p$ and $q$ respectively. Cyclic groups are abelian, and hence normal. Therefore $\langle x \rangle $ and $\langle y \rangle $ are non-trivial normal subgroups; hence $G$ is not simple. –  BobaFret Oct 28 '12 at 23:40
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@BobaFret No, $H \le G$ and $H$ abelian does $\bf{not}$ imply that $H \trianglelefteq G$. You might've been trying to use the fact that $G$ abelian implies that all subgroups are normal, but that doesn't apply here. –  Jackson Walters Nov 23 '12 at 4:04
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This CW answer intends to remove the question from the unanswered queue.


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

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