Find groups $G$, $H$ and a surjective homomorphism $\alpha: G\to H$ such that $\alpha(Z(G)) \neq Z(H)$ 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?
 A: There are plenty of examples:


*

*The determinant map $\text{GL}_n(\Bbb R)\rightarrow{\Bbb R}^\times$ when $n$ is even,

*The sign of the permutation map $S_n\rightarrow\{\pm 1\}$

*The orientation map $D_n\rightarrow\{\pm 1\}$ which to a plane isometry $\phi$ in the dihedral group $D_n$ assigns $1$ if and only if $\phi$ preserves orientation.


You should really work out these examples and convince yourself that they satisfy the condition requested.
A: No, your homomorphism is not surjective, and so your answer is not correct. In fact, every group of order 4 is abelian, so any surjective homomorphism between groups of order 4 will take the center (the whole group) to the center (the whole group).
A: As other people have already pointed out you're not going to get there with abelian groups. But hopefully you can get there with nonabelian groups. You might want to think about groups that don't have a lot of normal subgroups. In particular I would suggest looking at $S_5$, the symmetric group on $5$ elements.
A: Hint: The free group on at least $2$ letters has trivial center.
A: The easiest example is probably $\alpha: S_3\times C_2\rightarrow C_2$, where you kill the subgroup $A_3\times C_2$.
