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 $z\in \Omega_1(Z(G))\setminus\{1\}$ 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 each $z\in \Omega_1(Z(G))\setminus\{1\}$, 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.


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