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 $H<math>to<math>N$ 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 $\alpha (K)\leq K$ .

Related facts

Related subgroup properties satisfied by the center

Related subgroup properties not satisfied by the center

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 $\varphi (g)\neq e$ . But this means that $\alpha (e,g)\notin Z(G)$ , indicating that $\alpha (Z(G))$ is not a subgroup of $Z(G)$ .