Center not is injective endomorphism-invariant

Statement
It is possible for the center of a group to not be an injective endomorphism-invariant subgroup: in other words, there may be an injective endomorphism of the group that does not preserve the center.

Related facts

 * Center not is fully invariant
 * Center is strictly characteristic
 * Characteristic not implies injective endomorphism-invariant

Proof
Let $$A$$ be a nontrivial centerless group and $$B$$ be isomorphic to a nontrivial abelian subgroup $$C$$ of $$A$$, via a map $$\varphi$$. Define:

$$G := B \times A \times A \times A \times \dots$$.

In other words, $$G$$ is the external direct product involving one copy of $$B$$ and countably many copies of $$A$$.

The center of $$G$$ is the subgroup:

$$Z(G) = B \times \{ e \} \times \{ e \} \times \dots$$.

(Here, $$e$$ denotes the identity element).

Now consider the endomorphism $$\alpha$$ that maps the $$i^{th}$$ copy of $$A$$ isomorphically to the $$(i+1)^{th}$$ copy, and maps $$B$$ isomorphically to the subgroup $$C$$ in the first copy of $$A$$. $$\alpha$$ is clearly an injective endomorphism, and is explicitly described as:

$$\alpha(b,a_1,a_2, \dots, ) = (e,\alpha(b),a_1,a_2,\dots)$$.

(Here, $$e$$ denotes the identity element).

Also, $$\alpha(Z(G))$$ is not contained in $$Z(G)$$, so $$Z(G)$$ is not injective endomorphism-invariant.