# Index of subgroups in a finite solvable group, with trivial Frattini subgroup (Exercise 3B.12 from Finite Group Theory, by M. Isaacs)

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.

• Isn't $G=Alt(4)$, $M=C_2^2$, $H=C_2$ a counterexample? – verret Jun 1 at 7:47
• I think you're right. Since $G$ doesn't have a subgroup of order 6. – Mobius Jun 1 at 7:55
• 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 1 at 8:30

$$G=\mathrm{Alt}(4)$$, $$M=C_2^2$$, $$H=C_2$$ is a counterexample, as $$\mathrm{Alt}(4)$$ doesn't have a subgroup of order $$6$$. This seems like a mistake in the book.
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 $$|M:H|$$.
• 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 1 at 17:35