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Is it right to say for every (finite) abelian group $H$ there is non-abelian group $G$ such that $Z(G)=H$, where $Z(G)$ is the center of $G$?

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Yes, consider $G=H\times S_3$, which is a nonabelian group. We have $$ Z(G)=Z(H)\times Z(S_3)=Z(H)\times 1 \cong Z(H)=H, $$ since $Z(A\times B)=Z(A)\times Z(B)$ in general, and $S_3$ has trivial center.

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