- Show that the Quaternion group has only one element of order 2, namely \(-1\).
Solution
Write \[ Q_8=\{1,-1,i,-i,j,-j,k,-k\}, \] with \[ i^2=j^2=k^2=ijk=-1. \]
1. The element \(-1\) has order \(2\).
One has \((-1)^2=1\) and \(-1\ne1\).
2. Each of \(\pm i,\pm j,\pm k\) has order \(4\).
For \(u\in\{i,j,k\}\), \[ u^2=-1, \] so \(u^4=1\) but \(u^2\ne1\). Also \[ (-u)^2=u^2=-1, \] so \(-u\) likewise has order \(4\).
3. Hence \(-1\) is the unique element of order \(2\) in \(Q_8\).
The eight elements of \(Q_8\) are \(1,-1,\pm i,\pm j,\pm k\). The identity has order \(1\), \(-1\) has order \(2\) by 1, and all six remaining elements have order \(4\) by 2.