# no simple group of order $945$

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

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.