Final answer:
To prove that G/C is abelian, we use the property that the commutator subgroup C of G is normal in G and consists of elements of the form aba^{-1}b^{-1}. By definition, commutators commute with every element of G, and thus the factor group G/C, composed of cosets of C, is commutative. Therefore, G/C is an abelian group.
Step-by-step explanation:
To show that G/C is an abelian group, where C is the commutator subgroup of G, and assuming that every commutator can be expressed in the form aba-1b-1, we start by understanding the properties of a commutator subgroup. The commutator subgroup C of G is the subgroup generated by all elements of the form aba-1b-1, where a and b are elements of G. This subgroup is important because it measures the failure of G to be abelian.
By definition, the factor group G/C consists of cosets of the commutator subgroup C. To prove G/C is abelian, we need to show that the commutators commute with every element of G. Let x and y be arbitrary elements of G, then their cosets in G/C are xC and yC respectively.
We consider the product of two cosets xCyC. This is equal to (xy)C because C is normal in G (a property of being a commutator subgroup). The inverse of a coset x-1C is simply the coset of the inverse x-1C. We want to show that commutators [x,y] = xyx-1y-1 are in C by definition, and thus xCyC = yCxC. The result follows because the order of multiplication in G/C does not matter, thanks to the presence of all possible commutators in C.
Thus, for all x, y in G, we have xCyC = yCxC which demonstrates that G/C is an abelian group. The significance of this result is that by factoring out the commutator subgroup, we effectively 'cancel out' all the non-commutativity in G, resulting in an abelian group G/C.