Let $G$ be a group and $Z(G)$ its center. We can prove that $θ(Z(G))=Z(G)$ for every $θ \in{\rm Aut}(G)$
Can we find a group $G$ and an endomorphism $\varphi:G\to G$ s.t. $\varphi(Z(G))$ is not a subset of $Z(G)$?
My first thought is that the group $G$ cannot be abelian and that it can't have a trivial center.
My second thought was to take $D_4$ as $G$ whose center is a cyclic group but I couldn't find the right endomorphism.
Can you help?
Let $B$ be an Abelian subgroup of $A$ which is not central. Then there is a natural endomorphism on $A \times B$ given by $(a,b) \mapsto (b,1)$.
$B$ is not central in $A$ so there is $a' \in A$ and $b \in B$ which don't commute, so $$(a',1) \cdot (b,1) \ne (b,1) \cdot (a',1)$$ So the the image of $B$ isn't central.
$$(a',b') \dot (1,b) = (a',b'b) = (a',bb') = (a',b') \cdot (1,b)$$ As $B$ is commutative, so $B$ is central in $A \times B$