Exercise 4.3.13 (Finite groups which have exactly two conjugacy classes)

Proof. Suppose that the finite group $G$ has exactly two conjugacy classes. If $G$ is abelian, then the conjugacy classes have one element, so $|G|=2$ and $G\simeq Z_2$.

Suppose now that $G$ is not abelian. Let $Z=Z(G)$ be the center of $G$. If $|Z|\ge 3$, then $G$ has at least $3$ conjugation classes, and if $|Z|=2$, then $Z=\{a,b\}$, and $\{a\}$ and $\{b\}$ are two conjugacy classes. If $c$ is another element in $G$, then the conjugacy class of $c$ affords another conjugacy class. This is impossible, so in this case $G=Z$ is abelian, which contradicts the hypothesis, so

Since $G$ has exactly two conjugation classes, the class of $1$ and the class of some $r\ne 1$, the class equation has the form

where $d=|G:C_G(r)|$ divides $n=|G|$. Therefore $n-1\mid n=(n-1)+1$, so $n-1\mid 1$. This shows that $n=2$, so $G\simeq Z_2$, which contradicts the hypothesis “ $G$ is not abelian”.

In conclusion, the groups which have exactly two conjugacy classes are isomorphic to $Z_2$. □