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.

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