# 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) doesnotalways satisfy a particular subgroup property (i.e., retraction-invariant subgroup)

View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions

## Contents

## 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 is a group, the center of , denoted , is defined as:

.

### 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 of a group is termed a retraction if , i.e., for all .

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

A subgroup of a group is termed retraction-invariant if for any retraction of , we have .

## Related facts

### 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

## Proof

### A generic example

Let be a nontrivial centerless group, a nontrivial Abelian subgroup of , and a group isomorphic to via an isomorphism . Define:

.

Consider the normal subgroup:

and the complement to it given by:

.

Let be the retraction with kernel and image . In other words, takes any element of to the unique element of that is in the same coset of .

Note that , because is centerless and is Abelian. So, . For any , we have that and are in the same coset of . Thus, by definition, we have:

.

Since is an isomorphism and is nontrivial, there exists such that . But this means that , indicating that is not a subgroup of .