2
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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 $N = T^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.