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 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


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


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).