centralizer

Let $G$ be a group. The centralizer of an element $a\in G$ is defined to be the set

$$C(a)=\{x\in G\mid xa=ax\}$$

Observe that, by definition, $e\in C(a)$, and that if $x,y\in C(a)$, then $x{y}^{-1}a=x{y}^{-1}a(y{y}^{-1})=x{y}^{-1}ya{y}^{-1}=xa{y}^{-1}=ax{y}^{-1}$, so that $x{y}^{-1}\in C(a)$. Thus $C(a)$ is a subgroup of $G$. For $a\ne e$, 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 cosets of $C(a)$ and the conjugates of $a$.

Proof:
If $x,y\in G$ are in the same right coset, then $y=cx$ for some $c\in C(a)$. Thus $y^{-1}ay=x^{-1}{c}^{-1}acx=x^{-1}{c}^{-1}cax=x^{-1}ax$. Conversely, if $y^{-1}ay=x^{-1}ax$ then $x{y}^{-1}a=ax{y}^{-1}$ and $x{y}^{-1}\in C(a)$ giving $x,y$ are in the same right coset. Let $[a]$ denote the conjugacy class of $a$. It follows that $|[a]|=[G:C(a)]$ and $|[a]|\mid |G|$.

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

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

The groups $C(ga{g}^{-1})$ and $C(a)$, for any $g$, are isomorphic.

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	Normalizer
Related topic	GroupCentre
Related topic	ClassEquationTheorem