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Let $G$ be a simple group and let $n_p$ be the number of Sylow $p$-subgroups, $p$ prime. Show that $|G|$ divides $(n_p)!$ (factorial).

If i start off by assuming G is abelian then G is isomorphic to Z/pZ. So |G| = p, by Sylow's third theorem n_p is of the form n_p = 1+pk, k some integer. n_p must divide order of G witch implies k=0 and n_p = 1. But this would mean that the Sylow-p-subgroup is normal in G, contradicting the simplicity of G. Worse, |G|=p does not divide 1.

So i guess G is not abelian, not really sure how to continue from here.

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  • $\begingroup$ Maybe you mean $\prod n_p$ where the product ranges over the prime factors of $G$? $\endgroup$ – Pedro Tamaroff May 28 '14 at 22:57
  • $\begingroup$ It's written like this in my book. Maybe i misunderstand what you mean but wouldn't it be trivial in your case? $\endgroup$ – roslavets May 28 '14 at 23:01
  • $\begingroup$ As you observed the claim is wrong. $\endgroup$ – Pedro Tamaroff May 28 '14 at 23:05
  • $\begingroup$ How do we know it's wrong for nonabelian groups? $\endgroup$ – roslavets May 28 '14 at 23:10
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The number of Sylow $p$-subgroups equals the index $n=n_p$ of the normalizer $H$ of a Sylow $p$-subgroup. Now $G$ acts by left multiplication on the left cosets of $H$. If $K$ is the kernel of this action, then $G/K$ embeds homomorphically in $S_n$. But $G$ is simple so $K=1$, and hence $|G|$ divides $n!$.

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  • $\begingroup$ If $G=C_p$, $n_p=1$, but $G$ doesn't embbed into $S_1$. We need to assume $n_p>1$. $\endgroup$ – Pedro Tamaroff May 28 '14 at 23:17
  • $\begingroup$ If $n_p=1$ then your Sylow $p$-subgroup is normal. Thus the group is not simple. $\endgroup$ – Robert Wolfe May 28 '14 at 23:19
  • $\begingroup$ @Bryan True assuming $G$ is not $C_p$.If $G$ is a non-abelian $p$ group of size $>p$ it is not simple since $Z(G)$ is nontrivial normal subgroup of $G$. If $G$ is an abelian $p$-group it is trivially not simple for $n>1$, but it is simple for $n=1$, and it has a unique Sylow $p$-subgroup: itself! (Don't feel like you've been lied to, this is a very particular exception!) $\endgroup$ – Pedro Tamaroff May 28 '14 at 23:22
  • $\begingroup$ @Pedro's first comment: If $G$ is non-abelian simple, then this is automatically true. $\endgroup$ – Nicky Hekster May 28 '14 at 23:23
  • $\begingroup$ @NickyHekster Yes. $\endgroup$ – Pedro Tamaroff May 28 '14 at 23:33

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