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.