Center of a nonabelian p group can be found by taking the intersections of centers of two suitable maximal subgroups

Question

Let $P$ be a nonabelian $p$ group. Prove that there exist maximal subgroups $M$ and $N$ of $P$ such that

$$Z(P)=Z(M) \cap Z(N).$$

I found that $Z(P) \subseteq Z(M) \cap Z(N).$ I have no idea for the reverse inclusion.

Answer 1

$P/Z(P)$ is not cyclic, so (assuming that $P$ is finite) you can choose two distinct maximal subgroups $M,N$ containing $Z(P)$ and then $Z(P) \le Z(M) \cap Z(N)$.

For the other inclusion note that $M$ and $N$ generate $P$ and they both centralize $Z(M) \cap Z(N)$.