Is it possible for a finite group to have two isomorphic subgroups which is maximal and non-maximal

Let $G$ be a finite group. Is it possible that there exist two subgroups $H$ and $L$ of $G$ such that the following three conditions hold:

1. $H \cong L$
2. $H$ is maximal in $G$
3. $L$ is not maximal in $G$

Any example is welcomed. Thanks in advance.

Edit: This problem comes from computing unfactorizable morphisms in transporter category of group $G$. I searched for all subgroup of $S_5$ (See for the lattice of S5) containing a subgroup of order 2 and find out that if the 2-order subgroup is maximal so is all other 2-order subgroup. This is also true for subgroups containing 3-order subgroup. On the other hand, I searched the key world "maximal subgroup" in Dummits AA book, find no useful result to disprove the statement. I Also find no related result on the internet. So I conjecture that the answer is negative. But maybe a big finite group can be an example?

Solution 1:

Yes this is possible. There might be smaller examples, but one way to construct them is using twisted wreath products. The smallest example of this type is $G=A_5 \wr A_6$.

This has (at least) two conjugacy classes of subgroups isomorphic to $A_6$, one maximal and the other non-maximal.

The natural complement in the wreath product is not maximal in $G$. It normalizes the diagonal subgroup of the base group, resulting in a larger subgroup isomorphic to $A_5 \times A_6$.

However, there is another "twisted" complement of $G$ that is maximal. Roughly speaking, you twist it by making the stabilizer of the first block induce inner automorphisms of the first direct factor $A_5$ of the base group.

In my answer to this MO post, I gave this example as the smallest example of a primitive permutation group of twisted wreath product type inthe O'Nan-Scott Theorem.

$\textbf{Added Later}$. Here is a much smaller example, that you can check more easily on a computer. The Mathieu group $M_{12}$ has two conjugacy classes of subgroups isomorphic to ${\rm PSL}(2,11)$, one maximal, and the other non-maximal (contained in $M_{11}$).

$\textbf{Added Even Later}$. Thanks to MikkoKorhonen for pointing out that this question (asking specifically whether there are solvable examples) has already been discussed at some length in this MO post. I see from there that there is a smaller example than $M_{12}$. For a simple group $S$, the group $G = S \times S$ has two different types of subgroups isomorphic to $S$, the component subgroups, which are non-maximal, and the diagonal subgroups, which are maximal. The smallest such example is $A_5 \times A_5$ of order $3600$.