If $G$ and $H$ are groups and $f: G \to H$ is a surjective homomorphism, I can clearly see that $f(Z(G))\subseteq Z(H)$. However, I am unable to find examples where this subset relation is proper, i.e., $f(Z(G))\subsetneq Z(H)$.
The reason I am looking for such an example is because I want to prove that if $G$ is nilpotent and $f: G \to H$ is a surjective homomorphism then $H$ is nilpotent. I am using Dummit Foote to study this topic and the authors define a group to be nilpotent if its upper central series eventually becomes the entire group. If I could prove that the upper central series $Z_i (G)$ and $Z_i(H)$ satisfy that $f(Z_i(G))=Z_i(H)$ for each $i\in \mathbb Z _{\ge 0}$ then I will be done. However the case $i=1$ may not hold and this claim is apparently doomed.
I would appreciate if someone points me in the right direction to find a example of (nilpotent) groups $G$ and $H$ and a surjective group homomorphism $f:G\to H$ such that $f(Z(G))\subsetneq Z(H)$.
Best Answer
Take $A_5$, a simple non-abelian group. Consider $G=A_5\wr C_2$, the wreath product with the cyclic group of order $2$. Then the center of $G$ is trivial, but there is a surjective homomorphism $G\to C_2$ and the center of $C_2$ is $C_2$.
About nilpotent example, take $G=C_2\wr Q_8$, the wreath product of $C_2$ and the quaternion group. It is a 2-group, hence nilpotent. The center of it is inside $C_2^8$. Consider the natural homomorphism $G\to Q_8$, the image of the center of $G$ is trivial, but the center of $Q_8$ is not trivial.