Showing Sylow $p$-groups are Abelian if $n_p=2p+1$

Question

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?

Answer 1

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.

Answer 2

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.