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Let $G$ be a finite $p$-group of derived length $d$, which is not a Dedekind group. (i.e., possesses at least one non-normal subgroup). Let $G^{(d-1)}$ be the unique normal subgroup of $G$ of order $p$ ( so $Z(G)$ is cyclic). Let $H_1, H_2, \dots, H_t$ be the representatives of conjugacy classes of non-normal subgroups of $G$ and $t\geq 4$.

Is it possible that $G^{(d-1)}\not\leq H_1$ and $G^{(d-1)}\leq\cap_{i=2}^t H_i$?

(By an inspection with GAP, I could not find such a $p$-group. Observe that if $|H_1|\geq p^2$, then it contains a subgroup of order $p$, which is not normal in $G$ and does not contain $G^{(d-1)}$. Thus it seems that in this case the answer is negative. What does happen if $|H_1|=p$?)

Many thanks.

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You seem to be saying that a finite $p$-group which has a non-normal subgroup must have a cyclic center. A counterexample when $p=2$ is the group $G=D_8\times C_2$ of order 16. Its center is $C_2\times C_2$, so it has 3 normal subgroups of order $2$.

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