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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 1

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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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  • $\begingroup$ Thanks! It's lemma 2.7.13. I'll see where that takes me. $\endgroup$
    – user107736
    Commented Nov 10, 2013 at 9:03
  • $\begingroup$ Where is induction being used here? $\endgroup$
    – wasatar
    Commented Dec 4, 2019 at 21:09
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    $\begingroup$ @wasatar: In the case where $N \leq H$, we note that $H/N$ is a maximal subgroup of $G/N$ and apply induction on $G/N$. $\endgroup$ Commented Dec 4, 2019 at 21:35
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    $\begingroup$ @wasatar: Base case is $|G| = 1$. In the induction step, we use strong induction. That is, we assume that the result holds groups of order $< |G|$. $\endgroup$ Commented Dec 5, 2019 at 9:36
  • $\begingroup$ @MikkoKorhonen Oh I see now, since N is not trivial |G/N| < |G| and we can get the result. Thanks for clearing that up! $\endgroup$
    – wasatar
    Commented Dec 5, 2019 at 17:33

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