Suppose $G$ and $H$ are groups with $f:G\to H$ an isomorphism. Let $C_G$ be the center of $G$. Prove that $f(C_G) = \{ f(g) | g \in C_G\}$ is the center of $H$.
I have done this with my classmates and I did not understand it, and my teachers center discussion was a little brief. Can anyone show me the proof? Also my teacher says that this is a set equality proof, but there is no iff statement so why do we have to prove the converse?
Center of $H$ is the set $\{h\in H|\ \forall x\in H:hx=xh\}$. So the center of a group is the set of elements of the group that commute with all the elements of the group.
Let $h\in f(C_G)$. Then $h=f(g)$ for some $g\in C_G$. Since $g\in C_G,gx=xh$ for each $x\in G$. So $f(g)f(x)=f(gx)=f(xg)=f(x)f(g)$ for each $x\in G$. Now $hh^\prime=h^\prime h$ for each $h^\prime\in H$ (why?). So $h\in C_G.$ So $f(C_G)\subseteq C_H$.
What can you say of $f^{-1}:H\to G$? Do you see that $f^{-1}(C_H)\subseteq C_G$ and you can be done with the proof?