# Conjugacy Classes of Quaternion Group

## Theorem

Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group.

The conjugacy classes of $\Dic 2$ are:

$\set e, \set {a^2}, \set {a, a^3}, \set {b, a^2 b}, \set {a b, a^3 b}$

## Proof

From Center of Quaternion Group, we have:

$\map Z {\Dic 2} = \set {e, a^2}$

Thus from Conjugacy Class of Element of Center is Singleton, $\set e$ and $\set {a^2}$ are two of those conjugacy classes.

By inspection of the Cayley table:

$\begin{array}{r|rrrrrrrr} & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ \hline e & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ a & a & a^2 & a^3 & e & a b & a^2 b & a^3 b & b \\ a^2 & a^2 & a^3 & e & a & a^2 b & a^3 b & b & a b \\ a^3 & a^3 & e & a & a^2 & a^3 b & b & a b & a^2 b \\ b & b & a^3 b & a^2 b & a b & a^2 & a & e & a^3 \\ a b & a b & b & a^3 b & a^2 b & a^3 & a^2 & a & e \\ a^2 b & a^2 b & a b & b & a^3 b & e & a^3 & a^2 & a \\ a^3 b & a^3 b & a^2 b & a b & b & a & e & a^3 & a^2 \end{array}$

we investigate the remaining $6$ elements in turn, starting with $a$:

 $\ds a a a^{-1}$  $\ds$ $\ds = a$ $\ds a^2 a \paren {a^2}^{-1}$ $=$ $\ds a^2 a a^2 = a^5$ $\ds = a$ $\ds a^3 a \paren {a^3}^{-1}$ $=$ $\ds a^3 a a = a^5$ $\ds = a$ $\ds b a b^{-1}$ $=$ $\ds b a \paren {a^2 b} = b a^3 b$ $\ds = a^3$ $\ds \paren {a b} a \paren {a b}^{-1}$ $=$ $\ds \paren {a b} a \paren {a^3 b} = b a^3 b$ $\ds = a^3$ $\ds \paren {a^2 b} a \paren {a^2 b}^{-1}$ $=$ $\ds \paren {a^2 b} a b$ $\ds = a^3$ $\ds \paren {a^3 b} a \paren {a^3 b}^{-1}$ $=$ $\ds \paren {a^3 b} a \paren {a b} = \paren {a^2 b} \paren {a b}$ $\ds = a^3$

So we have a conjugacy class:

$\set {a, a^3}$

Investigating the remaining $4$ elements in turn, starting with $b$:

 $\ds a b a^{-1}$ $=$ $\ds a b a^3$ $\ds = a^2 b$ $\ds a^2 b \paren {a^2}^{-1}$ $=$ $\ds a^2 b a^2$ $\ds = b$ $\ds a^3 b \paren {a^3}^{-1}$ $=$ $\ds a^3 b a$ $\ds = a^2 b$ $\ds b b b^{-1}$  $\ds$ $\ds = b$ $\ds \paren {a b} b \paren {a b}^{-1}$ $=$ $\ds \paren {a b} b \paren {a^3 b} = \paren {a b} a^3$ $\ds = a^2 b$ $\ds \paren {a^2 b} b \paren {a^2 b}^{-1}$ $=$ $\ds \paren {a^2 b} b b = \paren {a^2 b} a^2$ $\ds = b$ $\ds \paren {a^3 b} b \paren {a^3 b}^{-1}$ $=$ $\ds \paren {a^3 b} b \paren {a b} = \paren {a^3 b} a$ $\ds = a^2 b$

So we have a conjugacy class:

$\set {b, a^2 b}$

Investigating the remaining $2$ elements, starting with $a b$:

 $\ds a \paren {a b} a^{-1}$ $=$ $\ds a \paren {a b} a^3 = a^2 b a^3$ $\ds = a^3 b$

We need go no further: the remaining elements $a b$ and $a^3 b$ are in the same conjugacy class:

$\set {a b, a^3 b}$

Hence the result.

$\blacksquare$