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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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    @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.2012-11-23

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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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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.

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Suppose without loss of generality that $p. By Cauchy's theorem, $G$ has a subgroup $H$ of order $q$. The index of this subgroup is $[G:H]=p$. Now recall that a subgroup of index the smallest prime divisor is normal.