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An old qual question on Sylow $p$-groups:

Assume that $p$ is an odd prime and $G$ is a finite simple group with exactly $2p+1$ Sylow $p$-groups. Prove that the Sylow $p$-groups of $G$ are abelian.

I am not sure which technique to apply to show that these groups are abelian. I know that the $2p+1$ Sylow $p$-subgroups are conjugate, but this is a statement about the entire sets rather than their individual elements and how they interact with each other. For any Sylow $p$-subgroup $P$ we have $\vert G/P\vert $=$\vert G\vert / p$.

I am not even sure if Sylow's theorem is needed since we are already given that $n_p=2p+1$. How can I proceed?

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  • $\begingroup$ Recall that the number of $p$-Sylow subgroup is equal to $|G:N_G(P)|$ where $P$ is a $p$-Sylow subgroup. $\endgroup$ – rafforaffo Jul 13 '15 at 10:15
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Hint: look at the normalizer of a Sylow $p$-subgroup. The index of this subgroup is $2p+1$. Then $G$ can be embedded in $A_{2p+1}$. Prove that $p^2$ is the highest power of $p$ dividing $(2p+1)!/2$. Use the fact that every group of order $p^2$ is abelian.

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  • $\begingroup$ Why can $G$ be embedded in $A_{2p+1}$? Where is the $p^2$ coming from? $\endgroup$ – The Substitute Jul 13 '15 at 10:10
  • $\begingroup$ $G$ acts on the left cosets of the normalizer by left multiplication (what I am using here is sometimes referred to as the $n!$ theorem). The kernel of this action is trivial, since $G$ is simple. $\endgroup$ – Nicky Hekster Jul 13 '15 at 10:15
  • $\begingroup$ In general, if a finite nonabelian simple group $G$ has a subgroup of index $n$, then $G$ embeds into $A_n$. $\endgroup$ – Derek Holt Jul 13 '15 at 10:17
  • $\begingroup$ @NickyHekster, is there an easier way than the $n!$ theorem to see that $G$ embeds into $A_{2p+1}$? $\endgroup$ – The Substitute Jul 13 '15 at 10:54
  • $\begingroup$ You wrote, this is a qual exam, study Theorem 1.1 of Isaacs' Finite Group Theory book for example. You will find it on Googel Books. $\endgroup$ – Nicky Hekster Jul 13 '15 at 12:16
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Let $A$ be the set of $2p+1$ Sylow $p$-subgroups in $G$. Let $G$ act on $A$ via conjugation. Consider the permutation representation $\phi:G\to S_A \le S_{2p+1}$. The kernel of $\phi$ is trivial since $G$ is simple. Hence $G$ embeds into $S_{2p+1}$. Therefore, orders of the elements of $A$ are at most $p^2$. Since groups of order $p,p^2$ are abelian, each element of $A$ is abelian.

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