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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?

Tian Vlašić
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JHJ
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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., http://groupprops.subwiki.org/wiki/Classification_of_groups_of_order_a_product_of_two_distinct_primes#Direct_proof_using_Sylow_results – 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 04:04

4 Answers4

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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.

Julian Kuelshammer
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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.

sandy
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Let's assume WLOG $p<q$. By Cauchy, there is a subgroup of order $q$. By contradiction, suppose there are more than one; say $H$ and $K$ two of them. Then, $HK\subseteq G$ and the cardinal of $HK$ is $q^2>pq$: contradiction. So, there is one subgroup of order $q$, only, which is then normal.

citadel
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Suppose without loss of generality that $p<q$. 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.

user162520
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