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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?2012-07-13
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    @Joe, Sorry it was a typing error2012-07-13

2 Answers 2

7

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.2012-07-14
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    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!2012-07-14
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    @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...).2012-08-21