Let $H\le G,G$ finite. If $M_H$ is a maximal subgroup of $H$, is it the case that $M_H = H \cap M$ for some $M$, a maximal subgroup of $G$?

Question

I have a question regarding finite groups and their subgroups. Specifically, let $G$ be a finite group, and let $H$ be a subgroup of $G$. I am interested in understanding whether the following statement holds true:

"If $M_H$ is a maximal subgroup of $H$, is it necessarily the case that $M_H = H \cap M$ for some $M$, a maximal subgroup of $G$?"

I believe that this statement is indeed true, and I have attempted to provide a proof for it. Here is my approach:

Let $h \in H \setminus M_H$. In order to establish the existence of $M$, a maximal subgroup of $G$, such that $M \cap H = M_H$, I propose considering $M$ as the maximal subgroup of $G$ that does not contain the subgroup $\langle h \rangle$ generated by $h$.

I would greatly appreciate any feedback, or hint. Thank you in advance for your time and assistance.

Answer 1

The answer to your question is no.

Let $G$ be cyclic of order $p^3$ for any prime $p$. Let $H$ be the unique maximal subgroup of order $p^2$ and let $M_H$ be the unique subgroup of order $p$. Then $G$ is a counterexample to your conjecture.

Answer 2

Let $G=Z_4$, and $H=\{0,2\}$. So $M_H=\{0\}$.

Now, $M=H$ is the non-trivial maximal subgroup of $G$, but $H\cap M=H\neq M_H$

Answer 3

As noted, the answer in general is no. In a cyclic group of prime power order, for example, the subgroups form a chain, so that given any two subgroups $H$ and $K$, either $H\leq K$ or $K\leq H$. In particular, there is a unique maximal subgroup $M$, and hence you will necessarily have $H\cap M=H$ for any proper subgroup $H$. This covers the examples given above by Mathfail and by Robert Shore.

More generally, if $G$ is a finite group, then $\Phi(G)$, the Frattini subgroup of $G$ is defined to be the intersection of all maximal subgroups of $G$. If $H\leq\Phi(G)$, then for every maximal subgroup we have $H\leq M$, hence $H\cap M = H$, so again you will not be able to reach your conclusion. (In the case of a nonabelian $p$-group $G$, we know that $\Phi(G)=G^p[G,G]$, which is nontrivial, so taking $H=\Phi(G)$ will give you counterexamples.) And if $M_H\cap \Phi(G)\neq H\cap \Phi(G)$, then you cannot find a maximal subgroup $M$ such that $M\cap H=M_H$: because if $h\in H\cap\Phi(G)$ with $h\notin M_H\cap\Phi(G)$, then $h\notin M_H$, but $h\in H\cap \Phi(G)\leq H\cap M$.

I did want to point out a few problems with your approach, for your future reference. First, note that it is incorrect to talk about "the maximal subgroup of $G$ that does not contain $h$": first, using the singular definite article the implies that such an object is unique; not only do you not know it is unique, you do not even know if it exists at all! You can certainly, in a finite group, find subgroups that are maximal with respect to not containing an element $h\neq e$, but these subgroups may (i) not be unique, so the use of "the" is unwarranted; and (ii) may fail to be maximal subgroups of $G$. Note that "maximal among subgroups of $G$ not containing $h$" does not imply "maximal among all proper subgroups of $G$.