A group 2-transitive with p+1 Sylow subgroups

Let $G$ be a simple group of Lie type and $2$-transitive such that the number of Sylow $p$-subgroups of $G$ is $p+1$. How we can prove that $G$ is isomorphic to $PSL(2,p)$? Thanks.

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The reference in the comment seems helpful: MO thread. But, I can't be sure as I don't know about these things. –  user21436 Aug 25 '12 at 7:20
@wentaway: Thanks, it was very useful, but still I have not proved why $G$ is isomorphic to $PSL(2,p)$ ? –  N K Aug 25 '12 at 8:00
As I explained in my comment in that thread, this follows, without using the classification of finite simple groups, from a result of Hering, Kantor and Seitz published in 1972. Is your problem that you do not understand how it follows, or that you wish to avoid using that result? –  Derek Holt Aug 25 '12 at 11:39
@DerekHolt: Yes my problem is that I do not understand how it follows. –  N K Aug 25 '12 at 11:57

1 Answer

Suppose that the group $G$ has $p+1$ Sylow $p$-subgroups for some prime $p$ and let $\Omega = {\rm Syl}_p(G)$ and $P \in \Omega$. Consider the action of $G$ on $\Omega$ by conjugation. Since $P$ cannot normalize any Sylow $p$-subgroup other than $P$ itself, $P$ fixes a unique point in this action, namely $P$ itself. The orbits of $P$ on $\Omega$ have length a power of $p$, so there must be exactly two orbits, of lengths $1$ and $p$. Since $G$ acts transitively (by Sylow's Theorem) and $P$ acts transitively on $\Omega \setminus \{ P \}$, $G$ acts 2-transitively. The stabilizer of $P$ in this action is the normalizer of $P$, so this stabilizer has $P$ as a normal subgroup, and in fact we must have $|P|=p$, so $P$ is a regular normal subgroup of the stabilizer of $P$ in its action on $\Omega \setminus \{ P \}$.

The result of Hering, Kantor and Seitz mentioned earlier is:

Let $G$ be a finite 2-transitive group on a set $\Omega$ and suppose that, for $\alpha \in \Omega$, $G_\alpha$ has a normal subgroup regular on $\Omega \setminus \{ \alpha\}$; then $G$ contains a normal subgroup $M$ and $M$ acts on $\Omega$ as one of the following groups in their usual 2-transitive representation: a sharply 2-transitive group, ${\rm PSL}(2,q)$, ${\rm Sz}(q)$, ${\rm PSU}(3,q)$ or a group of Ree type.

Since your group $G$ is simple, its action on $\Omega$ is faithful, and the hypotheses of this result apply. Simplicity implies that $M=G$. The groups ${\rm PSL}(2,p)$ are the only groups inthe conclusion with Sylow subgroups of order $p$.

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This is exactly what I was looking for! Thank you very much. –  N K Aug 25 '12 at 16:10