Suppose $T$ is a finite non-abelian simple group, $Inn(T^k) \leq G \leq Aut(T^k)$, and $G$ permutes the factors of $T^k$ transitively. Show that $G$ is a maximal subgroup in $T^k \rtimes G$, (where $G$ acts on $T^k$ as a subgroup of $Aut(T^k)$)
I can show that $T^k$ is a minimal normal subgroup in $T^k \rtimes G$. From here, I want to argue as follows: Suppose $G$ is not maximal. Then $\exists$ a subgroup of the form $N \rtimes G$, where $N \leq T^k$. Now I want to claim either that:
- $N \vartriangleleft T^k$ (in which case $N \cong T^j$ for some $j \leq k$, but the factors of $T^k$ are permuted transitively), or
- $N \vartriangleleft T^k \rtimes G$, which just implies that $N \vartriangleleft T^k$.
Obviously $N \vartriangleleft N \rtimes G$, but I don't know if this implies either $1$. or $2$., or if those statements are even true. I am new to semi-direct products so maybe there is a result that I am missing. Can someone point me in the right direction?
I solved it! If there is some subgroup $N \rtimes G$ in $T^k \rtimes G$, where $N \leq T^k$, then $N$ is invariant under the action of $G$ on $T^k$. In particular, since $Inn(T) \leq G$, $N \vartriangleleft T^k$. So $N = \prod_{j \in J}T$, for some $J \subset \{1,...,k\}$. As $G$ permutes the factors of $T^k$ transitively, we must have $N = T^k$.