Question

Use the class equation to show that if |G| ≥ 3, then G has at least 3 distinct conjugacy classes.

Answer 1

Let |G|=n, which is greater than or equals to 3. We consider two cases:

Case 1: Let G be abelian. In that case, G=Z(G). So all the conjugacy classes of G will contain single element. Thus, G contains as many conjugacy classes as the number of elements, i.e., n conjugacy classes.

Hence, the result holds when G is abelian.

Case 2: Let G be non-abelian. We know that Z(G) has at least one element. So,

$|Z(G)|\geq 1$

Since G is non-abelian, so there exists at least one $a\in G$ which does not commute with all elements of G. Let b be an element in G which does not commute with a. In that case,

$b\notin Z(G).$

We will show that b and a are not conjugate and hence they will have distinct conjugacy classes.

If possible, let a and b be conjugates. Then, there exists some g in G such that

$b=gag^{-1}$ .