All finite groups that have exactly 3 conjugacy classes .

Question

This question was asked in my abstract algebra quiz .

Determine all finite groups which have exactly 3 conjugacy classes .

I solved that $\mathbb{Z}/3$ is only group with conjugacy class 3.

While checking my argument by google search here :https://www.cefns.nau.edu/~falk/old_classes/511/extras/threeclass.html

I found that $S_{3}$ also has 3 conjugacy classes . But I have a question in the argument of the link .

Question :How does in the explanation given in the link author wrote :"m divides 1 + n and n divides 1 + m" in the 3rd line of the argument . I know the result"since the size of a conjugacy class equals the index of the centralizer of one of its elements)" which is given before the deduction but I dont know how to use it to deduce the result.

Can anyone please tell how to deduce the result ?

Answer 1

$|G|-m=1+n$. Since $m$ divides the order of the group and itself, it has to divide $1+n$. The same works for $n$.

Answer 2

according to class equation

$|G| = |C(x_1)| + |C(x_2)| + |C(x_3)|$

according to counting formula

$|G| = |Z(x_1)||C(x_1)| = |Z(x_2)||C(x_2)| = |Z(x_3)||C(x_3)|$

we have $1 = \frac{1}{|Z(x_1)|} + \frac{1}{|Z(x_2)|} + \frac{1}{|Z(x_3)|}$

case 1

let $|Z(x_1)| = 2, |Z(x_2)| = 3, |Z(x_3)| = 6$

accordingly, $|C(x_1)| = 3, |C(x_2)| = 2, |C(x_3)| = 1$

$G \cong S_3$

case 2

$|Z(x_1)| = |Z(x_2)| = |Z(x_3)| = 3$

accordingly, $|C(x_1)| = |C(x_2)| = |C(x_3)| = 1$

$G \cong C_3$

case 3

let $|Z(x_1)| = 2, |Z(x_2)| = |Z(x_3)| = 4$

accordingly, $|C(x_1)| = 2, |C(x_2)| = |C(x_3)| = 1$

$G \cong C_2 × C_2 \cong V$ (Klein four-group)

if the number of conjugacy classes is smaller than 3, it can be analyzed the same way