# Index of a maximal subgroup in a solvable group

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?

• At this moment I can only think of $A_4$, having Sylow $3$-subgroups of index $4$. Maybe think of $(\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2)\rtimes\Bbb Z_3$? – user441558 Oct 9 '17 at 14:49
• 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. – W4cc0 Oct 9 '17 at 14:51
• Edited previous comment. Not sure if the semi-direct product has a maximal Sylow $3$-subgroup or not... – user441558 Oct 9 '17 at 14:53
• I do not think so. The example has a non-trivial center, therefore one can always add it to the maximal subgroup. – W4cc0 Oct 9 '17 at 14:54
• 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 – 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 $G$.