Let $p$ be a prime and $n$ be a positive integer. Is it possible to find a finite solvable group $G$ with a maximal subgroup $M$ such that $|G:M|=p^n$?

If $n=1$, we can surely find it taking a group of order $pq$ with a normal Sylow $p$-subgroup. What if $n\geq 2$?

I can't even find examples of solvable groups with a maximal subgroup of cube prime index. Are there?

  • $\begingroup$ At this moment I can only think of $A_4$, having Sylow $3$-subgroups of index $4$. Maybe think of $(\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2)\rtimes\Bbb Z_3$? $\endgroup$ – user441558 Oct 9 '17 at 14:49
  • $\begingroup$ Yes, $A_4$ is one particular example. But I was wondering if there is some generic construction which shows what I wanted or maybe some theorem saying that is not possible. $\endgroup$ – W4cc0 Oct 9 '17 at 14:51
  • $\begingroup$ Edited previous comment. Not sure if the semi-direct product has a maximal Sylow $3$-subgroup or not... $\endgroup$ – user441558 Oct 9 '17 at 14:53
  • $\begingroup$ I do not think so. The example has a non-trivial center, therefore one can always add it to the maximal subgroup. $\endgroup$ – W4cc0 Oct 9 '17 at 14:54
  • $\begingroup$ Maybe $(\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2)\rtimes\Bbb Z_7$? Just a plain guess. Not sure if it has trivial center or not :P $\endgroup$ – user441558 Oct 9 '17 at 15:07

Let $p^n$ be a prime power and consider $G = \operatorname{AGL}(1, p^n)$, which is a semidirect product $\mathbb{F}_{p^n} \rtimes \mathbb{F}_{p^n}^*$. Then $\mathbb{F}_{p^n}^*$ is a maximal subgroup of index $p^n$ in $G$.

| cite | improve this answer | |

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.