the order of $Z(G)$

Question

Let $G$ be a non-abelian finite $p$-group and let $H$ be a proper subgroup of $G$ of order $p^2$. Suppose that $C_G(H) = H$. What is the order of $Z(G)$?

I just know that $|N_G(H)| = p^3$ and $1 \neq Z(G) \leq H$. Thanks!

Answer 1

The centralizer of any subgroup always contains the center of the group. So $C_G(H)=H$ tells you that $Z(G)\leq H$.

Now, that means that either $Z(G)$ has order $p$ or $p^2$. You know that $G$ has order at least $p^3$ (why?). If $Z(G)=H$, can the centralizer be exactly $H$?