Theorem : Let $G$ be a non-nilpotent group such that all the non-normal abelian subgroups of $G$ are cyclic. Then $G$ has cyclic center.
Proof. Suppose that $Z(G)$ is non-cyclic. since $G$ is non-nilpotent, then there exists some $P \in \mathcal{S} y \ell_{p}(G)$ such that $P \ntriangleleft G$. Then $Z(G)$ has a subgroup $\left\langle z_{1}, z_{2}\right\rangle$ isomorphic to $\mathbb{Z}_{q} \times \mathbb{Z}_{q},$ for some prime $(p \neq) q \in \pi(G) .$ since $\left\langle z_{1}, z_{2}, P\right\rangle \ntriangleleft G$ is abelian , then $\left\langle z_{1}, z_{2}, P\right\rangle$ is cyclic, which is a contradiction. Therefore $Z(G)$ is cyclic.
I can't understand why $Z(G)$ has a subgroup $\left\langle z_{1}, z_{2}\right\rangle$ isomorphic to $\mathbb{Z}_{q} \times \mathbb{Z}_{q} ? ,$ for some prime $(p \neq) q \in \pi(G) $ . why $p \neq q $ ? Why $\left\langle z_{1}, z_{2}, P\right\rangle \ntriangleleft G$? because $P \ntriangleleft G$ ? and how we can show $\left\langle z_{1}, z_{2}, P\right\rangle $ is abelian? I think proof of theorem is wrong.