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

$G$ is solvable, so we can consider the minimal normal subgroup $N$ in $G$.

I got the hint:

Apply induction to the quotient group $G/N$ and consider separately the two cases $N\le H$ and $N\nleq H$

But I still have no idea for this hint to follow. Is there anyone can give me more direction to proceed? any suggestion will be appreciated.Thanks for considering my request.

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1 Answer 1

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Hint:

Proof by induction. Remark that $N$ is not trivial so the cardinal of $G/H$ is strictly inferior to the cardinal of $G$. Let $p:G\rightarrow G/N$ the canonical projection.

  1. Suppose that $N\subset H$, $p(H)$ is a normal subgroup of $G/H$ and is maximal, so $[G:H]=[G/N:H/N]$ is prime, (see the reference)

if $N$ is not contained in $H$, remark that the cardinal of $N$ is prime, and $G$ is generated by $G$ and $n, n\in N$ we have $G=H\bigcup nH,...,n^{p-1}H$

Every minimal normal subgroup of a finite solvable group is elementary abelian

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  • $\begingroup$ Sorry, $G=H\bigcup nH,...,n^{p-1}H$ is the same meaning for $G=H\cup nH\,\cup\cdots\cup n^{p-1}H$ ? and I think you may write $G/N$ is strictly inferior to the cardinal of $G$, not $G/H$ ,in the very first line. $\endgroup$
    – user1992
    Dec 14, 2017 at 16:57
  • $\begingroup$ yes it is the same $\endgroup$ Dec 14, 2017 at 16:59
  • $\begingroup$ @TsemoAristide, you can't talk about $G/H$ being a group because $H$ might not be normal un $G$. And you should say "prime power", instead of only "prime". Anyway, your proof is nice. $\endgroup$ Nov 30, 2023 at 12:52

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