Conjugacy Classes of Quaternion Group

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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.

$\blacksquare$


Sources

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