Section 1.7 Conjugacy classes
ΒΆObjectives
You should be able to:
- Determine the conjugacy classes of a finite group.
Subsection 1.7.1 Definition
We have already discussed briefly the idea of equivalence classes. Given a group G, the idea is to define an equivalence relation, which we generally denote by β‘, between elements of the group, which is reflexive (aβ‘a), symmetric (if aβ‘b then bβ‘a) and transitive (if aβ‘b and bβ‘c then aβ‘c). Once we have an equivalence relation, we can define equivalence classes, which are subsets of elements of the group that are equivalent according to β‘. There is a particularly interesting equivalence relation, known as conjugation, that partitions any finite group into equivalence classes, known as conjugacy classes.Definition 1.7.1. Conjugacy classes.
We say that two elements a,bβG are conjugate if there exists an element gβG such that a=gbgβ1. This is an example of an equivalence relation. Then we can form equivalence classes, that is, classes of elements of G that are conjugate to each other. The set of all elements conjugate to one another is called a conjugacy class.
Subsection 1.7.2 Properties of conjugacy classes
It is clear that conjugacy classes are disjoint, since an element cannot be in two different conjugacy classes. So we can partition a group G into its conjugacy classes. It is also interesting to note that all elements in a given conjugacy class have the same order.Definition 1.7.2. Order of an element of a group.
We define the order of an element aβG to be the smallest integer n such that an=e.
Lemma 1.7.3.
All elements of a given conjugacy class have the same order.
Proof.
Suppose that \(a \in G\) has order \(n\text{.}\) If \(b\) is conjugate to \(a\text{,}\) that is \(b = g a g^{-1}\) for some \(g \in G\text{,}\) then
It follows that \(b^n = e\text{,}\) and hence the order of \(b\) is \(\leq n\text{.}\)
To finish the proof, we need to argue that there cannot be an integer \(m \lt n\) such that \(b^m= e\text{,}\) so that the order of \(b\) is precisely \(n\text{.}\) Suppose that there is such a \(m \lt n\text{.}\) Then
Multiplying by \(g^{-1}\) on the left and \(g\) on the right, we get that \(a^m = e\text{.}\) But \(n\) must be the smallest integer such that \(a^n = e\text{,}\) therefore this is a contradiction. It follows that the order of \(b\) is equal to \(n\text{.}\)
Lemma 1.7.4.
Each element in the centre of a group G constitutes a conjugacy class by itself. And hence the identity is always a conjugacy class by itself, and for an Abelian group G, each element forms its own distinct class.
Proof.
Let \(h \in Z(G)\) be an element of the centre of \(G\text{.}\) Then for all \(g \in G\text{,}\) we have \(h g = g h\text{,}\) and hence for all \(g \in G\text{,}\) \(g h g^{-1} = h\text{,}\) therefore the only element conjugate to \(h\) is \(h\) itself.
The two corollaries follow directly, since the identity is always an element of the centre of a group, and if \(G\) is abelian, then \(Z(G) = G\text{.}\)