# Index of maximal proper subgroup of a solvable group

This is problem 2.7.15 from Hungerford's Algebra:

If $H$ is a maximal proper subgroup of a finite solvable group $G$, then $[G:H]$ is a prime power.

If $G$ is abelian, then it's easy to show that $[G:H]$ is a prime power. I'm stuck on the non-abelian case. Any hints how to proceed?

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## 1 Answer

In a solvable group, you can show that a minimal normal subgroup is abelian of prime power order. I think this is a lemma in Hungerford that is proven in the section this problem is in.

Now here's a hint for your problem. Proceed by induction, and let $N \neq 1$ be a minimal normal subgroup of $G$. Consider the cases $N \leq H$ and $N \not\leq H$ separately.

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Thanks! It's lemma 2.7.13. I'll see where that takes me. – user107736 Nov 10 '13 at 9:03