Finite Group with 2 Conjugacy Classes has 2 Elements

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Let $G$ be a finite group.

Let $G$ have exactly $2$ conjugacy classes.

Then $G$ has exactly $2$ elements.


Let $G$ be of order $n$.

Let $G$ have exactly $2$ conjugacy classes.

Let $x \in G$ such that $x \ne e$.

Let $\conjclass x$ denote the conjugacy class of $x$.

From Identity of Group is in Singleton Conjugacy Class: $\conjclass e = \set e$ where $\conjclass e$ denotes the conjugacy class of $e$

The other elements of $G$ are in $\conjclass x$.


$\card {\conjclass x} = n - 1$

where $\card {\, \cdot \,}$ denotes the cardinality of a set.

We have that Size of Conjugacy Class is Index of Normalizer.

From Lagrange's Theorem, the index of any subgroup of $G$ is a divisor of $G$.


$\paren {n - 1} \divides n$

where $\divides$ denotes divisibility.

It follows that $n = 2$.

Hence the result by definition of order of group.