Simple subgroup of transitive group

Question

Suppose $Q \in Syl_n(G)$ and $Q$ is not normal in $G$, $Q$ is generated by an element of order $n$, where $G \leqslant S_n$ for prime $n$ and $G$ acts transitively on $\{1,...,n\}$. Define H to be a subgroup generated by all $g Q g^{-1}$ for $g \in G$. Clearly, $Q \leqslant H$ and clearly $H \trianglelefteq G$. Could you please suggest how to show that $H$ is noncommutative and simple?

Answer 1

Let $1 \ne N \unlhd H$. Since $H$ is transitive (because $Q$ is), the orbits of $N$ all have the same length and hence, since $n$ is prime, $N$ is transitive. So $n$ divides $|N|$, and $|H:N|$ is coprime to $n$. Hence $N$ contains all Sylow $n$-subgroups of $H$. But $H$ is generated by its Sylow $n$-subgroups, so $H=N$. The fact that $H$ is noncommutative follows from the fact that $Q$ is not normal in $G$ and hence $H$ has more than one Sylow $n$-subgroup. (If you didn't assume that $Q$ was not normal, then the proof that $H$ is simple is still valid, but it would be commutative.)