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.