## Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation)

**Point of post: **In this post we apply our methods of group actions to the particular group action of conjugation. We use this to derive the *class equation* and from this derive several key results the a corollary of which will be that every group of order with prime is abelian.

*Motivation*

In our last few posts we’ve spent a considerable amount of time developing the theory of group actions, orbits, orbit decompositions, etc. Now we’re going to put it to some real use. We’re going to consider a group acting on itself via conjugation. In other words, we’ll say that . From this we’ll be able to use the Orbit Decomposition Theorem to derive the *class equation* which states that if is a finite group then

where are the isotropy subgroups under the action of conjugation and is a set of representatives from each conjugacy class (i.e. the orbits under the action of conjugation) with more than one element. From this we’ll derive several important results including that if is a finite group with where is prime then is non-trivial. As a corollary of this we will be able to conclude that every group of order where is prime is abelian.

*Conjugation*

Let be a group and define the function

then is a -action on . Indeed

and

We call this action * acting on itself by conjugation*. For this particular action there are names for the orbits and isotropy subgroups. In particular, the isotropy subgroup of some is called the *normalizer of *and is denoted (or when the group under which we’re considering the action may be ambiguous) *.* The orbit of under the conjugation action is called the *centralizer of *and is denoted (or for the same reasons) and is called the *conjugacy class of *. We note the trivial fact that (where is the center of ) if and only if .

With this in mind we may now derive the class equation

**Theorem: ***Let be a finite group then*

*where is a set of representatives from the conjugacy classes which contain more than one element. If this sum is taken to be zero.*

**Proof: **By the Orbit Decomposition Theorem we know that

where is a set of representatives from each conjugacy class of . But, we can rewrite this as

By previous comment though

and thus if is the set of all such that then

Thus,

but, using the fact that we may conclude that

and the conclusion follows.

**References:**

1. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge: Cambridge UP, 1994. Print.

2. Lang, Serge. *Undergraduate Algebra*. 3rd. ed. Springer, 2010. Print.

3. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 200

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