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.



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

  • $\begingroup$ 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 '17 at 16:57
  • $\begingroup$ yes it is the same $\endgroup$ – Tsemo Aristide Dec 14 '17 at 16:59

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