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Let $q > p$ and $q, p$ be prime. Show that: No group $G$ with $|G|=pq$ is simple.

In class we didn't discuss the Sylow Theorem or the Chauchy Theorem yet. We only know about the Lagrange Theorem.

Or can you first tell me how the Lagrange Theorem says that there are subgroups of order $p$ and $q$.

I would be glad if you give me some hints to continue. Thank you for taking your time.

Chill2Macht
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maiT
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  • As we have discussed here, Lagrange does not give you the existence of subgroups of order $p$ and $q$. If you didn't discuss Cauchy's Theorem in class yet, you could read it at wikipedia. This would give you an advantage in class lateron, I think. – Dietrich Burde Oct 29 '16 at 08:48
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    Have a look at (http://math.stackexchange.com/q/222853). – Jean Marie Oct 29 '16 at 08:52

2 Answers2

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We can continue your question here, so that we have a subgroup $H$ of order $q$ in $G$. Recall that we have $q>p$. Because $[G: H ] = \frac{pq}{q}=p$, and $p$ is the smallest prime divisor of $G$, $H$ must be a normal subgroup of $G$, see here. Hence $G$ is not simple.

Community
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Dietrich Burde
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By Cauchy, there is a subgroup of order $q$, Say $H$. By contradiction, suppose there is another one, say $K$. Then, $HK\subseteq G$ and $|HK|=\frac{|H||K|}{|H\cap K|}=$ $\frac{q^2}{1}=$ $q^2>pq$: contradiction. So, $H(=K)$ is unique, and hence normal in $G$.

citadel
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