# Proper subgroups which have the same minimal subgroups(revision)

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.

## 1 Answer

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 $C_{17} \times C_{17}$.

• Recall that I said "Suppose also that every $H_j$ is a non-maximal subgroup". But in your example the subgroup $P$ is a maximal subgroup and also contains all minimal subgroups of order 17 as same as other $H_j$s. Am I right? – H.Shahsavari Sep 4 '16 at 21:25
• P is not a maximal subgroup. – Geoff Robinson Sep 4 '16 at 22:40
• Index $G/P \cong C_9$, so $P$ is not maximal. The reason this example is a as large as it is, is that $G/P$ had to have non-prime order, and I couldn't find examples with $|G/P|$ even. But it is possible that there are smaller examples. – Derek Holt Sep 5 '16 at 7:33
• Dear professors Holt and Robinson. Thank you for your helpful answers. – H.Shahsavari Sep 8 '16 at 11:11
• If you believe that this is indeed a counterexample, then you should accept the answer. – Derek Holt Sep 8 '16 at 12:54