With reference to John B Fraleigh's Abstract Algebra 7th ed page 164, To study a nonabelian group we abelianize it by forming its quotient with commotator subgroup $C$ (which is generated by commutators of $G$) and that quotient is abelian.We have made all commutators equal to identity in $G/C$.
The thing I cannot understand is what about other elements of $G$? How do the elements in cosets other than commutator subgroup have commutative property as we do computations with representatives.How do we get $ab = ba$ in $G/C$? Explaination through examples will be appreciated. Thanks for your valuable comments.
Note that for any $a,b \in G$ we have $aba^{-1}b^{-1}\in C$, applying the quotient map $\pi: G \to G/C$, $g \mapsto [g] := gC$ to this equation, we have $$ [a][b][a]^{-1}[b^{-1}] = 1 \iff [a][b]=[b][a] . $$ Hence, $G/C$ is abelian.