Conjugacy Classes of the Quaternion Group, Q8

Let $Q_8 = \{ 1, -1, i, \bar{i}, j, \bar{j}, k, \bar{k} \}$ be The Quaternion Group, Q8 (see link for the group operation table). We will described all of the conjugacy classes of $Q_8$.

We first look at $Z(Q_8)$. Recall that $ij = k \neq \bar{k} = \neq ji$, $jk = i \neq \bar{i} =\neq kj$, and $ki = j \neq \bar{j} = ik$. Thus $i, j, k \not \in Z(Q_8)$. Similarly $\bar{i}j \neq j \bar{i}$, $\bar{j}k \neq k\bar{j}$, and $\bar{k}i \neq i\bar{k}$, so $\bar{i}, \bar{j}, \bar{k} \not \in Z(Q_8)$. We see that only $1, -1 \in Z(Q_8)$.

Recall that $g \in G$ has a trivial conjugacy class if and only if $g \in Z(G)$. Thus we see that the conjugacy classes of $1$ and $-1$ are trivial, that is:

(1)

\begin{align} \quad [1] &= \{ 1 \} \\ \quad [-1] &= \{ -1 \} \end{align}

Now look at the element $i \in Q_8$. We see that:

(2)

\begin{align} \quad iii^{-1} &= i \\ \quad \bar{i}i\bar{i}^{-1} &= \bar{i}i\bar{i} = \bar{i} \\ \quad jij^{-1} &= ji\bar{j} = \bar{k}\bar{j} = \bar{i} \\ \quad \bar{j}i\bar{j}^{-1} &= \bar{j}ij = kj = i \\ \quad kik^{-1} &= ki\bar{k} = j \bar{k} = \bar{i} \\ \quad \bar{k}i\bar{k}^{-1} &= \bar{k}ik = \bar{j}k = \bar{i} \end{align}

Therefore the conjugacy class of $i$ is $[i] = \{ i, \bar{i} \}$. It can similarly be shown that the conjugacy classes of $[j]$ and $[k]$ are as expected, $[j] = \{j, \bar{j} \}$ and $[k] = \{k, \bar{k} \}$. Thus, $Q_8$ is partioned into its conjugacy classes by:

(3)

\begin{align} \quad Q_8 &= [1] \cup [-1] \cup [i] \cup [j] \cup [k] = \{ 1 \} \cup \{ -1 \} \cup \{ i, \bar{i} \} \cup \{ j, \bar{j} \} \cup \{k, \bar{k} \} \end{align}

Moreover, The Class Equation for $Q_8$ is:

(4)

\begin{align} \quad 8 = 2 + (2 + 2 + 2) \end{align}