centralizer


Let G be a group. The centralizerMathworldPlanetmath of an element aG is defined to be the set

C(a)={xGxa=ax}

Observe that, by definition, eC(a), and that if x,yC(a), then xy-1a=xy-1a(yy-1)=xy-1yay-1=xay-1=axy-1, so that xy-1C(a). Thus C(a) is a subgroupMathworldPlanetmathPlanetmath of G. For ae, the subgroup is non-trivial, containing at least {e,a}.

To illustrate an application of this concept we prove the following lemma.

Lemma:
There exists a bijection between the right cosetsMathworldPlanetmath of C(a) and the conjugatesPlanetmathPlanetmath of a.

Proof:
If x,yG are in the same right coset, then y=cx for some cC(a). Thus y-1ay=x-1c-1acx=x-1c-1cax=x-1ax. Conversely, if y-1ay=x-1ax then xy-1a=axy-1 and xy-1C(a) giving x,y are in the same right coset. Let [a] denote the conjugacy classMathworldPlanetmath of a. It follows that |[a]|=[G:C(a)] and |[a]||G|.

We remark that aZ(G)C(a)=G|[a]|=1, where Z(G) denotes the center of G.

Now let G be a p-group, i.e. a finite groupMathworldPlanetmath of order pn, where p is a prime and n is a positive integer. Let z=|Z(G)|. Summing over elements in distinct conjugacy classes, we have pn=|[a]|=z+aZ(G)|[a]| since the center consists precisely of the conjugacy classes of cardinality 1. But |[a]|pn, so pz. However, Z(G) is certainly non-empty, so we conclude that every p-group has a non-trivial center.

The groups C(gag-1) and C(a), for any g, are isomorphicPlanetmathPlanetmathPlanetmathPlanetmath.

Title centralizer
Canonical name Centralizer
Date of creation 2013-03-22 12:35:01
Last modified on 2013-03-22 12:35:01
Owner drini (3)
Last modified by drini (3)
Numerical id 14
Author drini (3)
Entry type Definition
Classification msc 20-00
Synonym centraliser
Related topic NormalizerMathworldPlanetmath
Related topic GroupCentre
Related topic ClassEquationTheorem