Center not is retraction-invariant

Statement with symbols
The center of a group need not be a retraction-invariant subgroup: in other words, there may be a retraction of the whole group under which the center is not mapped to within itself.

Center
The center of a group is defined as the set of all those elements of the group that commute with every element of the group. In symbols, if $$G$$ is a group, the center of $$G$$, denoted $$Z(G)$$, is defined as:

$$Z(G) = \{ z \in G \mid gz = zg \ \forall \ g \in G \}$$.

Retraction-invariant subgroup
A retraction of a group is an idempotent endomorphism from the group to itself. In other words, an endomorphism $$\alpha$$ of a group $$G$$ is termed a retraction if $$\alpha^2 = \alpha$$, i.e., $$\alpha(\alpha(g)) = \alpha(g)$$ for all $$g \in G$$.

Retractions arise as follows: Pick a complemented normal subgroup $$N$$ of $$G$$ and a permutable complement H to $$N$$ in $$G$$. In other words, $$NH = G$$ and $$N \cap H$$ is trivial. Then, we can define a retraction $$\alpha$$ of $$G$$ that sends every element to the unique element of $$H$$ that lies in the same coset of $$N$$.

A subgroup $$K$$ of a group $$G$$ is termed retraction-invariant if for any retraction $$\alpha$$ of $$G$$, we have $$\alpha(K) \le K$$.

Related subgroup properties satisfied by the center

 * Center is direct projection-invariant
 * Center is strictly characteristic
 * Center is quasiautomorphism-invariant
 * Center is characteristic

Related subgroup properties not satisfied by the center

 * Center not is fully characteristic
 * Center not is I-characteristic
 * Center not is intermediately characteristic

A generic example
Let $$A$$ be a nontrivial centerless group, $$B$$ a nontrivial Abelian subgroup of $$A$$, and $$C$$ a group isomorphic to $$B$$ via an isomorphism $$\varphi:C \to B$$. Define:

$$G = A \times C$$.

Consider the normal subgroup:

$$N = A \times \{ e \}$$

and the complement to it given by:

$$H = \{ (\varphi(g),g) \mid g \in C \}$$.

Let $$\alpha:G \to G$$ be the retraction with kernel $$N$$ and image $$H$$. In other words, $$\alpha$$ takes any element of $$G$$ to the unique element of $$H$$ that is in the same coset of $$N$$.

Note that $$Z(G) = \{ e \} \times C$$, because $$A$$ is centerless and $$C$$ is Abelian. So, $$\alpha(Z(G)) = \alpha( \{ e \} \times C)$$. For any $$g \in C$$, we have that $$(\varphi(g),g)$$ and $$(e,g)$$ are in the same coset of $$N$$. Thus, by definition, we have:

$$\alpha(e,g) = (\varphi(g),g)$$.

Since $$\varphi$$ is an isomorphism and $$C$$ is nontrivial, there exists $$g \in C$$ such that $$\varphi(g) \ne e$$. But this means that $$\alpha(e,g) \notin Z(G)$$, indicating that $$\alpha(Z(G))$$ is not a subgroup of $$Z(G)$$.