Introduction to group theory/Socks and shoes proof

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Proof

Let G be a group and let a,b∈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−1a−1 by definition of inverse.

Q.E.D.
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