What are the Conjugacy Classes of Q8

The conjugacy classes of Q8 are {1}, {−1}, {i,−i}, {j,−j} and {k,−k}. Recall that

\begin{equation*} Q_8 = \{1,-1,i,-i,j,-j,k,-k\} \end{equation*}

where

(Q_8, \cdot)

is a group with product operation,

1 \cdot a = a \cdot 1 = 1

(-1) \cdot a = a \cdot (-1) = -1

for all

a \in Q_8

(-1)\cdot(-1) = 1

and

\begin{align*} i \cdot i &= j \cdot j = k \cdot k = -1 \\ i \cdot j &= k, \quad i \cdot k = -j, \\ j \cdot i &= -k, \quad j \cdot k = i, \\ k \cdot i &= j, \quad k \cdot j = -j \end{align*}

Recall that to find the conjugacy classes, for each element of

Q_8

say

a

, we have to find all the element

\sigma a\sigma^{-1}

for all

\sigma \in Q_8

. This corresponds to a conjugacy class.

Proof. It is easily to see for 1 and -1 that we have the conjugacy classes

\{-1\}

and

\{1\}

. Now can do the rest:

a = i

\begin{align*} 1 \cdot i \cdot 1 &= i \\ (-1) \cdot i \cdot (-1) &= i \\ i\cdot i \cdot i &= -i \\ (-i)\cdot i \cdot (-i) &= (-i)\cdot 1 = -i \\ j \cdot i \cdot j &= j\cdot k = i \\ (-j)\cdot i \cdot (-j) &= (-j)\cdot(-k) = i \\ k \cdot i \cdot k &= k \cdot (-j) = i \\ (-k)\cdot i \cdot (-k) &= (-k)\cdot j = i \\ \end{align*}

a = -i

\begin{align*} 1 \cdot (-i) \cdot 1 &= -i \\ (-1) \cdot (-i) \cdot (-1) &= -i \\ i\cdot (-i) \cdot i &= i \\ (-i)\cdot (-i) \cdot (-i) &= (-i)\cdot (-1) = i \\ j \cdot (-i) \cdot j &= j\cdot (-k) = -i \\ (-j)\cdot (-i) \cdot (-j) &= (-j)\cdot k = -i \\ k \cdot (-i) \cdot k &= k \cdot j = -i \\ (-k)\cdot (-i) \cdot (-k) &= (-k)\cdot (-j) = -i \\ \end{align*}

So we have the conjugacy class

\{i,-i\}

a = j

\begin{align*} 1 \cdot j \cdot 1 &= j \\ (-1) \cdot j \cdot (-1) &= j \\ i\cdot j \cdot i &= i \cdot (-k) = j \\ (-i)\cdot j \cdot (-i) &= (-i)\cdot k = j \\ j \cdot j \cdot j &= -j \\ (-j)\cdot j \cdot (-j) &= -j \\ k \cdot j \cdot k &= k \cdot i = j \\ (-k)\cdot j \cdot (-k) &= (-k)\cdot (-i) = j \\ \end{align*}

a = -j

\begin{align*} 1 \cdot (-j) \cdot 1 &= -j \\ (-1) \cdot (-j) \cdot (-1) &= -j \\ i\cdot (-j) \cdot i &= i \cdot k = -j \\ (-i)\cdot (-j) \cdot (-i) &= (-i)\cdot (-k) = -j \\ j \cdot (-j) \cdot j &= j \\ (-j)\cdot (-j) \cdot (-j) &= j \\ k \cdot (-j) \cdot k &= k \cdot (-i) = -j \\ (-k)\cdot (-j) \cdot (-k) &= (-k)\cdot i = -j \\ \end{align*}

So we have the conjugacy class

\{j,-j\}

a = k

\begin{align*} 1 \cdot k \cdot 1 &= k \\ (-1) \cdot k \cdot (-1) &= k \\ i\cdot k \cdot i &= i \cdot j = k \\ (-i)\cdot k \cdot (-i) &= (-i)\cdot (-j) = k \\ j \cdot k \cdot j &= j \cdot (-i) = k \\ (-j)\cdot k \cdot (-j) &= (-j) \cdot i = k \\ k \cdot k \cdot k &= -k \\ (-k)\cdot k \cdot (-k) &= -k \\ \end{align*}

a = -k

\begin{align*} 1 \cdot (-k) \cdot 1 &= -k \\ (-1) \cdot (-k) \cdot (-1) &= -k \\ i\cdot (-k) \cdot i &= i \cdot (-j) = -k \\ (-i)\cdot (-k) \cdot (-i) &= (-i)\cdot j = -k \\ j \cdot (-k) \cdot j &= j \cdot i = -k \\ (-j)\cdot (-k) \cdot (-j) &= (-j) \cdot (-i) = -k \\ k \cdot (-k) \cdot k &= k \\ (-k)\cdot (-k) \cdot (-k) &= k \\ \end{align*}

So we have the conjugacy class

\{k,-k\}

. Conclusion: {1}, {−1}, {i,−i}, {j,−j} and {k,−k} are the conjugacy classes of the quaternion group Q8.