Finite groups of class 2 e 3.

Let G be a finite group. We want to classify all finite groups with exactly two and three conjugacy classes. The number of conjugacy classes of G is called the class of G and is denoted by Cl(G). In the following if x ∈ G the conjugacy class of x is denoted by Cl(x). Then let G be a finite group and suppose first that G has exactly two conjugacy classes, let’s say Cl(a) is Cl(b). Since G is a disjoint union of its conjugacy classes we can assume that a=e (since in general if y ∈ Cl(x) then Cl(y)=Cl(x)) so called C_G(b) the centralizer of b in G we have:

|G | = 1 + |Cl(b)| = 1+ [G : C_G(b)] =

1 + |G|/|C_G(b)|

and so:

|G|-1=|G|/|C_G(b)| —> |G|-1 divide |G| —> |G|=2 —> G≅ℤ₂.

Similarly if G has order 2 it has exactly two conjugacy classes. Now suppose that G has order 3 and let Cl(a),Cl(b) and Cl(c) be the conjugacy classes of G. Clearly if G is abelian it is ℤ₃ which has three conjugacy classes. So let’s assume that G is not abelian. As before we can assume that a=e. Then:

|G|=1+|G|/|C_G(b)| + |G|/|C_G(c)|

and so:

1=1/|G| + 1/|C_G(b)| + 1/|C_G(c)|.

Without losing generality we can assume that |C_G(b)|≤ |C_G(c)|. Now since |G|≥3 the only possibility is that |C_G(b)| ≤ 3 . If |C_G(b)|=2 then |C_G(c)|≥3 so |G|≤6 but G is non-abelian so G≅S₃ which in fact has exactly three conjugacy classes. If |C_G(b)|=3 then |C_G(c)|=3 so |G|=3 absurd since G is non-abelian. To summarize:

Theorem:

Let G be a finite group. Then:

Cl(G)=2 <—> G≅ℤ₂

Cl(G)=3 <—>G≅ℤ₃ o G≅S₃.