Let $ G$ be a finite non-Dedekind $ p$ -group with non-cyclic center, where $ p$ is an odd prime.

By $ [\langle x\rangle]_G=\{g^{-1}\langle x\rangle g\ |\ g\in G\},$ I mean the conjugacy class of the subgroup $ \langle x\rangle$ of $ G$ and $ U=\{[\langle x\rangle]_G\ |\ \langle x\rangle\ntrianglelefteq G\}$ is the set of all conjugacy classes of non-normal cyclic subgroups of $ G$ . Also $ \Omega_1(G)$ denotes the subgroup generated by all elements of $ G$ of order at most $ p$ . Let $ z\in \Omega_1(Z(G))$ and consider the following subsets of $ U$ \begin{eqnarray*} &&U_1=\{[\langle x\rangle]_G\in U\ |\ \langle z\rangle\not\leq \langle x\rangle, \ \ [\langle x\rangle]_G=\{ \langle xz^i\rangle\ |\ 0\leq i\leq p-1\}\}\ &&U_2 =\{[\langle x\rangle]_G\in U\ |\ \langle z\rangle\not\leq \langle x\rangle,\ \ [\langle x\rangle]_G\cap\{ \langle xz^i\rangle\ |\ 1\leq i\leq p-1\}=\phi\} \end{eqnarray*}

I want to give an affirmative or negative answer to the follwing question

Question: If for all $ z\in \Omega_1(Z(G))$ , we have $ U_1=U_2=\phi$ , then is $ \mbox{exp}(Z(G))=p$ and $ G^{\prime}$ an elementary abelian $ p$ -group?

I could not find any counter example but in the following minor case I know the answer to the first part of the Question is yes,

Let all non-normal subgroups of $ G$ be cyclic and let there exists an element $ x\in \Omega_1(G)\setminus Z(G)$ . In the contrary case, if there exists $ z\in Z(G)$ such that $ o(z)>p$ , then $ \Omega_1(\langle x,z\rangle)\cong C_p\times C_p$ and by assumption $ \Omega_1(\langle x,z\rangle)\lhd G$ . Now since $ \langle x\rangle\leq \Omega_1(\langle x,z\rangle)$ , it follows that $ [\langle x\rangle]_G\in U_1$ , which is a contradiction.

Any help will be greatly appreciated.