Question:
Find groups $G$ and $H$ and a surjective homomorphism $\alpha: G \to H$ such that $\alpha(Z(G)) \neq Z(H)$
My answer:
Let $G$ and $H$ both be cyclic groups of order 4.
Define $\alpha: G \to H$ such that $g \in G \to I_h \in H$.
So now the center of $G$, $Z(G)$ is all of $G$ and it is getting mapped by $\alpha$ to $I_h$. $Z(H) =$ all of $H = \{I_h, a, a^2, a^3\}$ which is not equal to $\{I_h\}$.
Does that look correct?
* EDIT: Revised answer, does this look correct? *
Let $G$ be the dihedral group or order 8:
$\{I_G, a, a^2, a^3, x, ax, a^2x, a^3x \}$
Let $H$ by the cyclic group of order $8$.
$\{I_G, b, b^2, b^3, b^4, b^5, b^6, b^7 \}$
Define a surjective homomorphism $\alpha:G \to H$ as
$I_G \to I_H$
$ a\to b$
$ a^2\to b^2$
$ a^3\to b^3$
$ x\to b^4$
$ ax\to b^5$
$ a^2x\to b^6$
$ a^3x\to b^7$
Now $Z(G) = \{I_G, a^2\}$ and $Z(H) = H$.
Therefore, $\alpha(Z(G)) = \{I_H, b^2\} \neq H = Z(H)$
Does that look correct?