I asked a similar question yesterday. But I found out that I should make some major changes to it. So I decided to ask a new version of my previous question.


Prove or find a counterexample to the following statement.


Let $G$ be a finite group with at least two minimal subgroups and $I\_1, I\_2, ... ,I\_n$ be a family of minimal subgroups of $G$. Suppose that $H\_1, H\_2, ... ,H\_m$ are all the subgroups of $G$ such that for each $i,j$, $I\_i\leq H\_j$ and there is not any minimal subgroup other than $I\_i$s contained in $H\_j$ for all $1\leq j\leq m$ (i.e. $H\_j$s have the same minimal subgroups and there is not any other subgroup such that it's minimal subgroups are as same as $H\_j$s). Suppose also that every $H\_j$ is a **non-maximal** subgroup. Then there exists at least one minimal subgroup say $L$ such that $L\nsubseteq H\_1\cup H\_2\cup ... \cup H\_m$ and $\langle H\_j,L\rangle \neq \langle H\_k,L \rangle$ for all $j\neq k$. Furthermore $\langle H\_1,L\rangle, \langle H\_2,L\rangle, ... \langle H\_m,L\rangle$ are proper subgroups of $G$ which have the same minimal subgroups.



Proper subgroups which have the same minimal subgroups(revision)
There are so many conditions that it is hard to find examples satisfying the hypotheses, which makes it harder to find counterexamples, but I am sure they exist.


I believe there is a counterexample that is an extension of $P=C\_{17^2} \times C\_{17^2}$ by $C\_9$ with fixed point free action of $C\_9$ on $P$. For the $H\_i$ take all subgroups of $P$ that contain $C\_{17} \times C\_{17}$. The only minimal subgroups of $G$ outside of $P$ are of order $3$, and there is a unique conjugacy class of these. Any such subgroup $L$ of order $3$ generates $\langle P,L \rangle$ with all $H\_i$ except for $C\_{17} \times C\_{17}$.


