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:

  1. $N \vartriangleleft T^k$ (in which case $N \cong T^j$ for some $j \leq k$, but the factors of $T^k$ are permuted transitively), or
  2. $N \vartriangleleft T^k \rtimes G$, 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?

  • $\begingroup$ 1. is a bit imprecise. To formulate it accurately, you should write $T^k=T^K$, with $K$ a finite set on which $G$ acts transitively. Then the "efficient" formulation in 1. is that the normal subgroups of $T^K$ are (and are not just isomorphic) the $T^I$ for $I$ ranging over subsets of $K$. $\endgroup$ – YCor Mar 21 '19 at 16:14
  • $\begingroup$ $G$ is not maximal in $T^k\rtimes G$. Indeed, choose any nontrivial proper subgroup $C$ of $T$; then $G\subset C^k\rtimes G\subset T^k\rtimes G$ are proper inclusions. $\endgroup$ – YCor Mar 21 '19 at 16:15
  • $\begingroup$ @YCor To consider $C^k \rtimes G$ as a semi-direct product, you need that $C^k$ is invariant under the action of $G$ on $T^k$ $\endgroup$ – vxnture Mar 21 '19 at 16:18
  • $\begingroup$ Yes, the above applies only when $G$ acts just permuting the factors, preserving coordinates. It's a counterexample to the assertion that $G$ is maximal. It's unclear from your first paragraph if this is an assertion, or an assumption. "I want to argue as follows": argue for what? $\endgroup$ – YCor Mar 21 '19 at 16:22
  • $\begingroup$ I've edited it to make it clear. I want to show that $G$ is a maximal subgroup in $T^k \rtimes G$, where the underlying action of $G$ is as a group of automorphisms of $T^k$. But I think I see the solution now! $\endgroup$ – vxnture Mar 21 '19 at 16:27

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 $N = T^k$.


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