Groups of order 8

Let \(G\) have order \(8\).

If \(G\) is abelian, the classification of finite abelian groups gives exactly \[ \mathbb Z/8,\qquad \mathbb Z/4\times\mathbb Z/2,\qquad (\mathbb Z/2)^3. \]

Assume \(G\) is nonabelian. Then \(G\) has an element \(r\) of order \(4\): if every nonidentity element had order \(2\), then for all \(x,y\), \[ (xy)^{-1}=xy, \] while also \((xy)^{-1}=y^{-1}x^{-1}=yx\), so \(xy=yx\). Thus \(H=\langle r\rangle\) has order \(4\) and index \(2\), hence is normal. Choose \(s\notin H\). Then \(G=H\sqcup sH\), and conjugation by \(s\) induces an automorphism of \(H\cong C_4\). Since \(G\) is nonabelian, this automorphism is nontrivial, so \[ srs^{-1}=r^{-1}. \] Also \(s^2\in H\). Since \(s\) commutes with its own square, \(s^2\) cannot be \(r\) or \(r^3\): either equality would force \(s\) to commute with \(r\), contradicting \(srs^{-1}=r^{-1}\ne r\). Hence \[ s^2\in\{1,r^2\}. \]

If \(s^2=1\), then \[ G\cong \langle r,s\mid r^4=s^2=1,\ srs^{-1}=r^{-1}\rangle, \] the dihedral group of order \(8\). If \(s^2=r^2\), then \[ G\cong \langle r,s\mid r^4=1,\ s^2=r^2,\ srs^{-1}=r^{-1}\rangle, \] which is \(Q_8\).

Hence the five groups of order \(8\) are \[ \mathbb Z/8,\quad \mathbb Z/4\times\mathbb Z/2,\quad (\mathbb Z/2)^3,\quad D_8,\quad Q_8, \] where \(D_8\) denotes the dihedral group of order \(8\).