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)
View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions
Statement with symbols
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:
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 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 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 .