# Conjugacy classes of non-normal subgroups of a finite $p$-group

Let $G$ be a finite $p$-group of derived length $d$ and nilpotency class $c$. Suppose that $G$ is not a Dedekind group (i.e., possesses at least one non-normal subgroup). Suppose that $G^{(d-1)}$ has order p and is the unique normal subgroup of order $p$. Let $s$ be the number of conjugacy classes of non-trivial subgroups of $G$ that do not contain $G^{(d-1)}$ (thanks to the assumptions on $G$, they are automatically non-normal).

If $G$ has at least $4$ conjugacy classes of non-normal subgroups, then is it true that $s\geq 3$?

By an inspection with GAP, I could check this inequality for such $p$-groups of small order in GAP library which I checked. Does there exist a method to prove this inequality? Any comment or answer will be appreciated!

• "of small order": maybe you want to say up to which order you checked. – YCor Mar 31 '18 at 13:21
• Dear YCor, thank's for your mention. I editted the question. – sebastian Mar 31 '18 at 13:28
• Your condition that $G$ has a unique normal subgroup of order $p$ is equivalent to requiring that $Z(G)$ is cyclic. – Geoff Robinson Mar 31 '18 at 18:07