Let $G$ be a finite $2$-group of class 2 (that is $G^{\prime}\leq Z(G)$) and $Z(G)$ be cyclic. We could verify that $G^{\prime}=\langle [a,b]\rangle$ for some $a, b \in G$. If $H=\langle a,b\rangle$, then by a well-known argument we have $G=HC_G(H)$ and $G^{\prime}=H^{\prime}$. Let $|G^{\prime}|=|H^{\prime}|=2^n$ for some integer $n>0$. Thus we have ${\rm exp}(\frac{G}{Z(G)})={\rm exp}(\frac{H}{Z(H)})=2^n$.
How could we show that $Z(H)=\langle a^{2^n}, b^{2^n}, [a,b]\rangle$?
(Since ${\rm exp}(\frac{H}{Z(H)})=2^n$ and $G=HC_G(H)$,
$\langle a^{2^n}, b^{2^n}, [a,b]\rangle\leq Z(H)\leq Z(G)$. In particular $Z(H)$ is cyclic.
I could not show the converse of the first inclusion)
Any answer will be greatly appreciated!
Notations: By $C_G(H)$ I mean $C_G(H):=\{g\in G | gh=hg, \forall h\in H\}$ and ${\rm exp}(G)$ denotes the exponent of the group $G$.