Center not is retraction-invariant

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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

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