Blog

What are the Conjugacy Classes of Q8

The conjugacy classes of Q8 are {1}, {−1}, {i,−i}, {j,−j} and {k,−k}. Recall that
Q8={1,−1,i,−i,j,−j,k,−k}\begin{equation*} Q_8 = \{1,-1,i,-i,j,-j,k,-k\} \end{equation*}
where (Q8,⋅)(Q_8, \cdot) is a group with product operation, 1⋅a=a⋅1=11 \cdot a = a \cdot 1 = 1, (−1)⋅a=a⋅(−1)=−1(-1) \cdot a = a \cdot (-1) = -1 for all a∈Q8a \in Q_8, (−1)⋅(−1)=1(-1)\cdot(-1) = 1 and
i⋅i=j⋅j=k⋅k=−1i⋅j=k,i⋅k=−j,j⋅i=−k,j⋅k=i,k⋅i=j,k⋅j=−j\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 Q8Q_8 say aa, we have to find all the element σaσ−1\sigma a\sigma^{-1} for all σ∈Q8\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}\{-1\} and {1}\{1\}. Now can do the rest: a=ia = i:
1⋅i⋅1=i(−1)⋅i⋅(−1)=ii⋅i⋅i=−i(−i)⋅i⋅(−i)=(−i)⋅1=−ij⋅i⋅j=j⋅k=i(−j)⋅i⋅(−j)=(−j)⋅(−k)=ik⋅i⋅k=k⋅(−j)=i(−k)⋅i⋅(−k)=(−k)⋅j=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=−ia = -i:
1⋅(−i)⋅1=−i(−1)⋅(−i)⋅(−1)=−ii⋅(−i)⋅i=i(−i)⋅(−i)⋅(−i)=(−i)⋅(−1)=ij⋅(−i)⋅j=j⋅(−k)=−i(−j)⋅(−i)⋅(−j)=(−j)⋅k=−ik⋅(−i)⋅k=k⋅j=−i(−k)⋅(−i)⋅(−k)=(−k)⋅(−j)=−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}\{i,-i\}.

a=ja = j:
1⋅j⋅1=j(−1)⋅j⋅(−1)=ji⋅j⋅i=i⋅(−k)=j(−i)⋅j⋅(−i)=(−i)⋅k=jj⋅j⋅j=−j(−j)⋅j⋅(−j)=−jk⋅j⋅k=k⋅i=j(−k)⋅j⋅(−k)=(−k)⋅(−i)=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=−ja = -j:
1⋅(−j)⋅1=−j(−1)⋅(−j)⋅(−1)=−ji⋅(−j)⋅i=i⋅k=−j(−i)⋅(−j)⋅(−i)=(−i)⋅(−k)=−jj⋅(−j)⋅j=j(−j)⋅(−j)⋅(−j)=jk⋅(−j)⋅k=k⋅(−i)=−j(−k)⋅(−j)⋅(−k)=(−k)⋅i=−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}\{j,-j\}.

a=ka = k:
1⋅k⋅1=k(−1)⋅k⋅(−1)=ki⋅k⋅i=i⋅j=k(−i)⋅k⋅(−i)=(−i)⋅(−j)=kj⋅k⋅j=j⋅(−i)=k(−j)⋅k⋅(−j)=(−j)⋅i=kk⋅k⋅k=−k(−k)⋅k⋅(−k)=−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=−ka = -k:
1⋅(−k)⋅1=−k(−1)⋅(−k)⋅(−1)=−ki⋅(−k)⋅i=i⋅(−j)=−k(−i)⋅(−k)⋅(−i)=(−i)⋅j=−kj⋅(−k)⋅j=j⋅i=−k(−j)⋅(−k)⋅(−j)=(−j)⋅(−i)=−kk⋅(−k)⋅k=k(−k)⋅(−k)⋅(−k)=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}\{k,-k\}. Conclusion: {1}, {−1}, {i,−i}, {j,−j} and {k,−k} are the conjugacy classes of the quaternion group Q8.
Tags: ,

Leave a Reply