Example of proper image of center of group under epimorphism.

Question

Let $f:G \to H$ be a group epimorphism and let $Z(G)$ be a center of a group $G$. We already know that $f(Z(G)) \subseteq Z(H)$. Is there any small example showing $f(Z(G))$ is a proper subgroup of $Z(H)$?

Answer 1

Consider $f:GL_2(\Bbb{R})\rightarrow\Bbb{R}^\times$ defined by $$f(A)=|A|$$ Note that $Z(GL_2(\Bbb{R}))=\{kI_2:k\in\Bbb{R}^\times\}$ while $Z(\Bbb{R}^\times)=\Bbb{R}^\times$.
Next, $|kI_2|=k^2$ hence there does not exist $A\in Z(GL_2(\Bbb{R}))$ such that $f(A)=-1$.
Thus $$f(Z(GL_2(\Bbb{R})))<Z(\Bbb{R}^\times)$$

Answer 2

I just found out answer. Take $G = S_{3}, H= C_{2}$, and $f$ is canonical morphism by identifying $H\cong S_{3}/A_{3}$. Then we have $Z(G) = 0, Z(H)=C_{2}$.

Answer 3

The claim is not true. Take for instance $G = C_4$, $H = S_4$ and $f\colon G \to H$ the homomorphism sending $G$ to a $4$-cycle in $H$. Then the center of $G$ is all of $G$, but the center of $H$ is trivial.