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I need to show that there are no simple groups of order $945$.

I've tried the regular method using the Sylow theorems.

$$|G|=945=3^3\cdot5\cdot7 $$

If $G$ is simple then there should be 7 Sylow-3 groups ; 21 Sylow-5 groups and 15 Sylow-7 groups. Even if they would all intersect trivially, there will still be no contradiction.

Any ideas?

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Why are you trying to find a contradiction by assuming $G$ is not simple? – Joe Johnson 126 Jul 13 '12 at 20:32
@Joe, Sorry it was a typing error – catch22 Jul 13 '12 at 20:34
up vote 6 down vote accepted

If as you say there are 7 Sylow 3-subgroups, then the normalizer $N$ of one of these subgroups has index 7. If $G$ is simple, then there is an injective permutation representation $G \hookrightarrow S_7$, and so $|G| = 945$ must divide $7!$, but this is not the case.

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I think every simple group has even order.

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Every non-abelian finite simple group has even order. – Zev Chonoles Jul 14 '12 at 2:24
Why does every student first learning group theory seriously make that mistake?Really,I had a fellow graduate student who made that mistake the first time he was TAing the undergraduate abstract algebra course! – Mathemagician1234 Jul 14 '12 at 2:26
@ZevChonoles: Couldn't that be further specified to „of nonprime order”? Or am I missing something here? I'm not questioning the downvotes (Feit-Thompson is obviously not what this question was asking about...). – tomasz Aug 21 '12 at 15:34

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