7
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Isn't $G=Alt(4)$, $M=C_2^2$, $H=C_2$ a counterexample? $\endgroup$ – verret Jun 1 at 7:47
  • $\begingroup$ I think you're right. Since $G$ doesn't have a subgroup of order 6. $\endgroup$ – David Burrell Jun 1 at 7:55
  • 1
    $\begingroup$ 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 $N$. $\endgroup$ – Derek Holt Jun 1 at 8:30
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wow! Thanks for posting. Awesome that you stopped in. $\endgroup$ – amWhy Jun 1 at 16:34
  • 1
    $\begingroup$ 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! $\endgroup$ – Derek Holt Jun 1 at 17:35
  • 1
    $\begingroup$ Marty, great that you chimed in! Does there exist a list of errata? $\endgroup$ – Nicky Hekster Jun 1 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.