Introduction to group theory/Socks and shoes proof

Proof

Let $G$ be a group and let $a,b\in G$. Then $(a*b)*(b^{-1}*a^{-1})=a*(b*b^{-1})*a^{-1}=a*e*a^{-1}=a*a^{-1}=e$. Also $(b^{-1}*a^{-1})*(a*b)=b^{-1}*(a^{-1}*a)*b=b^{-1}*e*b=b^{-1}*b=e$. Thus $(ab{)}^{-1}=b^{-1}{a}^{-1}$ by definition of inverse.

Q.E.D.