Let G {\displaystyle G} be a group and let g , a , b ∈ G {\displaystyle g,a,b\in G} such that a ∗ g = e = g ∗ a {\displaystyle a*g=e=g*a} and b ∗ g = e = g ∗ b {\displaystyle b*g=e=g*b} . Then substituting we obtain a ∗ g = b ∗ g {\displaystyle a*g=b*g} . By right cancellation a = b {\displaystyle a=b} .
