Index of maximal proper subgroup of a solvable group

Question

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?

score 4 · Accepted Answer · answered Nov 10 '13 at 08:59

4

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.

answered Nov 10 '13 at 08:59

Mikko Korhonen

26,156

Thanks! It's lemma 2.7.13. I'll see where that takes me. – user107736 Nov 10 '13 at 09:03
Where is induction being used here? – wasatar Dec 04 '19 at 21:09
1

@wasatar: In the case where $N \leq H$, we note that $H/N$ is a maximal subgroup of $G/N$ and apply induction on $G/N$. – Mikko Korhonen Dec 04 '19 at 21:35
1

@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 $< |G|$. – Mikko Korhonen Dec 05 '19 at 09:36
@MikkoKorhonen Oh I see now, since N is not trivial |G/N| < |G| and we can get the result. Thanks for clearing that up! – wasatar Dec 05 '19 at 17:33

Index of maximal proper subgroup of a solvable group

1 Answers1

Linked