Prove that if N is a normal subgroup of G then the multiplication on G/N given by aN*bN = abN is well defined?

Question

Prove that if N is a normal subgroup of G then the multiplication on G/N given by aN*bN = abN is well defined

Answer 1

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Since N is normal bNb^-1 = N ==> aNbN = a(bNb^-1)(bN) = abN(b^-1b)N = abN1N = abN.