# Center not is retraction-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., retraction-invariant subgroup)
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## Statement

### 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.

## Definitions used

### Center

Further information: 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

Further information: Retraction,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 $H[itex] to [itex]N$ 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 $\alpha(K) \le K$.

## Proof

### 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 $Z(G)$.