Center not is injective endomorphism-invariant

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) does not always satisfy a particular subgroup property (i.e., injective endomorphism-invariant subgroup)
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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.