# Proof that if $G$ permutes the factors of $T^k$ transitively, $G$ is maximal in $T^k \rtimes G$

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?

• 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 $K$. – YCor Mar 21 '19 at 16:14
• $G$ 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. – YCor Mar 21 '19 at 16:15
• @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 $T^k$ – vxnture Mar 21 '19 at 16:18
• 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? – YCor Mar 21 '19 at 16:22
• 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! – 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 $$N = T^k$$.