What is the center of the quaternion group \(Q_8\)? Show that \(Q_8/Z(Q_8)\) is abelian.
Solution
Write \[ Q_8=\{\pm1,\pm i,\pm j,\pm k\}, \] with \(i^2=j^2=k^2=ijk=-1\).
1. The center of \(Q_8\) is\[ Z(Q_8)=\{\pm1\}. \]
The elements \(1\) and \(-1\) commute with every element of \(Q_8\), so \(\{\pm1\}\subseteq Z(Q_8)\).
On the other hand, \[ ij=k\qquad\text{while}\qquad ji=-k, \] so neither \(i\) nor \(-i\) is central. Cyclically permuting \(i,j,k\) shows that none of \(\pm j,\pm k\) is central either. Thus no further elements lie in the center.
2. The quotient \(Q_8/Z(Q_8)\) has four elements, each of order dividing \(2\).
By 1, the quotient has order \(8/2=4\). Its cosets are \[ Z(Q_8),\quad iZ(Q_8),\quad jZ(Q_8),\quad kZ(Q_8). \] Moreover, \[ (iZ(Q_8))^2=i^2Z(Q_8)=(-1)Z(Q_8)=Z(Q_8), \] and similarly for the \(j\)- and \(k\)-cosets.
3. Hence\[ Q_8/Z(Q_8)\cong \ZZ_2\oplus\ZZ_2, \]in particular the quotient is abelian.
By 2 the quotient has order \(4\) and no element of order \(4\), so it is the Klein four group rather than the cyclic group of order \(4\). Therefore it is isomorphic to \(\ZZ_2\oplus\ZZ_2\) and is abelian.