|
Let  be a group. The centralizer of an element  is defined to be the set
Observe that, by definition,
 , and that if
 , then
 , so that
 . Thus  is a subgroup of  . For  , the subgroup is non-trivial, containing at least  .
To illustrate an application of this concept we prove the following lemma.
Lemma:
There exists a bijection between the right cosets of  and the conjugates of  .
Proof:
If  are in the same right coset, then  for some
 . Thus
 . Conversely, if
 then
 and
 giving  are in the same right coset. Let ![$ [a]$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img22.png "$ [a]$") denote the conjugacy class of  . It follows that
![$ \vert[a]\vert = [G : C(a)]$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img24.png "$ \vert[a]\vert = [G : C(a)]$") and
![$ \vert[a]\vert \mid \vert G\vert$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img25.png "$ \vert[a]\vert \mid \vert G\vert$") .
We remark that
![$ a \in Z(G) \iff C(a) = G \iff \vert[a]\vert = 1$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img26.png "$ a \in Z(G) \iff C(a) = G \iff \vert[a]\vert = 1$") , where  denotes the center of  .
Now let  be a  -group, i.e. a finite group of order  , where  is a prime and  is a positive integer. Let
 . Summing over elements in distinct conjugacy classes, we have
![$ p^n = \sum{\vert[a]\vert} = z + \sum_{a \notin Z(G)}{\vert[a]\vert}$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img35.png "$ p^n = \sum{\vert[a]\vert} = z + \sum_{a \notin Z(G)}{\vert[a]\vert}$") since the center consists precisely of the conjugacy classes of cardinality  . But
![$ \vert[a]\vert \mid p^n$](http://images.planetmath.org:8080/cache/objects/2833/l2h/img37.png "$ \vert[a]\vert \mid p^n$") , so  . However,  is certainly non-empty, so we conclude that every  -group has a non-trivial center.
The groups
 and  , for any  , are isomorphic.
|
