Table of contents |

2 Properties 3 Conjugacy class equation 4 Conjugacy of subgroups 5 Conjugacy of general subsets |

Suppose *G* is a group. Two elements *a* and *b* of *G* are called **conjugate** iff there exists an element *g* in *G* with *g ^{-1}ag* =

**Cl(**= {*a*)*x*in*G*: there exists*g*in*G*such that*x*=*g*}^{-1}ag

If *G* is abelian, then *g ^{-1}ag* =

If two elements *a* and *b* of *G* belong to the same conjugacy class (i.e. if they are conjugate), then they have the same order. More generally, every statement about *a* can be translated into a statement about *b*=*g ^{-1}ag*, because the map φ(

An element *a* of *G* lies in the center Z(*G*) of *G* if and only if its conjugacy class has only one element, *a* itself. More generally, if C_{G}(*a*) denotes the *centralizer* of *a* in *G*, i.e. the subgroup consisting of all elements *g* such that *ga* = *ag*, then the index [*G* : C_{G}(*a*)] is equal to the number of elements in the conjugacy class of *a* (this is a special case of a result for conjugacy of subsets given later, as can be seen by letting *S* = {*a*} in the equation below).

If *G* is a finite group, then the previous paragraphs, together with the Theorem of Lagrange, implies that the number of elements in every conjugacy class divides the order of *G*.

Furthermore, for any group *G*, we can define a representative set *S* = {*x*_{i}} by picking one element from each equivalence class of *G* which has more than one element. *S* then has the property that *G* is the disjoint union of Z(*G*) and the conjugacy classes Cl(*x*_{i}) of the elements of *S*. One can then formulate the following important **class equation**:

- |
*G*| = |Z(*G*)| + ∑_{i}[*G*:*H*_{i}]

As an example of the usefulness of the class equation, consider a group *G* with order *p*^{n}, where *p* is a prime number and *n* > 0. Since the order of any subgroup of *G* must divide the order of *G*, it follows that each *H*_{i} also has order some power of *p*^{( ki )}. But then the class equation requires that |*G*| = *p*^{n} = |Z(*G*)| + ∑_{i} (*p*^{( ki )}). From this we see that *p* must divide |Z(*G*)|, so |Z(*G*)| > 1, and therefore we have the result: *every finite p-group has a non-trivial center*.

One can also look at conjugation as a group action on the set of all subgroups of a group G. They fall into orbitss, called again **conjugacy classes** of subgroups. The stabilizer of a given subgroup H for this action is its normalizer.

More generally, given any subset *S* of *G* (*S* now not necessarily a subgroup), we define a subset *T* of *G* to be conjugate to *S* if and only if there exists some *g* in *G* such that *T* = *g ^{-1}Sg* (the notation

A frequently used theorem is that, given any subset *S* of *G*, the index of N(*S*) (the normalizer of *S*) in *G* equals the order of Cl(*S*):

- |Cl(
*S*)| = [*G*: N(*S*)]