# centralizer

Let $G$ be a group. The 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 $xy^{-1}a=xy^{-1}a(yy^{-1})=xy^{-1}yay^{-1}=xay^{-1}=axy^{-1}$, so that $xy^{-1}\in C(a)$. Thus $C(a)$ is a subgroup   of $G$. For $a\neq e$, the subgroup is non-trivial, containing at least $\{e,a\}$.

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

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 $xy^{-1}a=axy^{-1}$ and $xy^{-1}\in C(a)$ 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 $C(gag^{-1})$ and $C(a)$, for any $g$, are isomorphic    .

Title centralizer Centralizer 2013-03-22 12:35:01 2013-03-22 12:35:01 drini (3) drini (3) 14 drini (3) Definition msc 20-00 centraliser Normalizer  GroupCentre ClassEquationTheorem