Let $G$ be a simple group which acts on $\Omega$. Let $\alpha \in \Omega$ such that $|O(\alpha)|=p$. Prove the order of $p$-sylow subgroup is $p$.

Question

Let $G$ be a finite simple group which acts on $\Omega$.

Let $\alpha \in \Omega$ such that $|O(\alpha)|=p$, ($O$ is the orbit of $\alpha$, $p$ is a prime number).

Prove the order of $p$-sylow subgroup is $p$.

My attempt:

$|O(\alpha)|=p$ hence $\phi:G\to S_p$ is an homomorphism.

Using first isomorphism theorem $G/N \cong S_p.$

Since $G$ is simple, $N=\{1\} \implies G\cong S_p$.

$|G|=|S_p|=p! \implies $using Lagrange's theorem we conclude the order of $p$-sylow subgroup is $p$.

Is my solution correct?

Thanks!

Answer 1

$G$ acts (transitively) on every orbit of the parent action. Since $G$ is simple, any such induced action is either trivial or faithful. If $|O(\alpha)|=p$, then the faithful $G$-action on $O(\alpha)$ induces an embedding of $G$ into $S_p$, and hence $|G|$ divides $p!$. Now, $p$ divides $|G|$ (orbit-stabilizer theorem), and no higher powers of $p$ can divide $p!$.