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Let G be a finite solvable group, and assume that $\Phi(G) = 1$ where $\Phi(G)$ denotes the Frattini subgroup of G. Let M be a maximal subgroup of G, and suppose that $H \subseteq M$. Show that $G$ has a subgroup with index equal to $|M:H|$.

This is question 3B.12 from Finite Group Theory, by M. Isaacs.

Here is my approach so far. I am completely stuck and would welcome any hints or ideas.

Suppose otherwise. Among all of the counter examples choose $G$ of minimum order. Since $G$ is a counterexample it must be the case that $|G| > 1$. Since $G$ is a counter example there is a maximum subgroup $M$ and a subgroup $H \subset M$, such that every subgroup of $G$ does not have the same index as $|M:H|$. So it must be the case that $H$ is properly contained within $M$.

This is where I get stuck. I want to use a minimal normal subgroup $N$ of $G$ which exists. But my argument devolves into a series of cases about whether or not $N$ intersects $H$ and/or $M$ non-trivially.

I do know that $G$ must have a non-normal maximal subgroup, since if they all were normal then it would be nilpotent and since G is finite this implies supersolvable, then $G$ would have a subgroup for any divisor of its order. Since $\Phi(G)=1$ is the intersection of all the maximal subgroups of $G$ I suspect this should help but I'm not sure where to go from here.

verret
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DGB
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    Isn't $G=Alt(4)$, $M=C_2^2$, $H=C_2$ a counterexample? – verret Jun 01 '20 at 07:47
  • I think you're right. Since $G$ doesn't have a subgroup of order 6. – DGB Jun 01 '20 at 07:55
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    I just checked that the question has been correctly cited from the book, so I guess this must be a rare mistake in the book! Note that if $G$ had a minimal normal subgroup $N$ not contained in $M$, then we would have $N \cap M=1$ and $G=NM$, so $NH$ would be the required subgroup. But in general there is no such $N$. – Derek Holt Jun 01 '20 at 08:30

2 Answers2

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Yes, I regret that Problem 3B.12 of my group theory book is wrong. It should be replaced by the following:

Let $H \subseteq M \subseteq G$, where $M$ is a maximal subgroup of a solvable group $G$, and assume that the core of $M$ in $G$ is trivial. Show that $G$ has a subgroup with index equal to $|M:H|$.

I. M. Isaacs

Marty Isaacs
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    Wow! Thanks for posting. Awesome that you stopped in. – amWhy Jun 01 '20 at 16:34
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    So, with the additional assumption that $M$ is core-free in $G$, a minimal normal subgroup $N$ is elementary abelian and not contained in $M$, so $NM=G$, $N \cap M = 1$, and $NH$ is the required subgroup. I won't give any more details! – Derek Holt Jun 01 '20 at 17:35
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    Marty, great that you chimed in! Does there exist a list of errata? – Nicky Hekster Jun 01 '20 at 20:21
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$G=\mathrm{Alt}(4)$, $M=C_2^2$, $H=C_2$ is a counterexample, as $\mathrm{Alt}(4)$ doesn't have a subgroup of order $6$. This seems like a mistake in the book.

verret
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