This is part of Exercise 2.7.9 of F. M. Goodman's "Algebra: Abstract and Concrete".
Definition 1: The commutator subgroup $C$ of a group $G$ is the subgroup generated by all elements of the form $xyx^{-1}y^{-1}$ for $x, y\in G$.
The Question: Show that the commutator subgroup $C$ of a group $G$ is normal and that $G/C$ is abelian.
My Attempt:
Let $g\in G$. Then $g(xyx^{-1}y^{-1})g^{-1}=gxy(gyx)^{-1}$, but I don't know where to go from here. What I'm trying to do is write $g(xyx^{-1}y^{-1})g^{-1}$ as an element of $C$.
Please help :)
We have $(gxg^{-1})^{-1}=gx^{-1}g^{-1}$ and similarly for $gyg^{-1}$, so that $$\begin{align} g(x yx^{-1}y^{-1})g^{-1}&=gx\cdot (g^{-1}g)\cdot y\cdot (g^{-1}g)\cdot x^{-1}\cdot (g^{-1}g)\cdot y^{-1}g^{-1} \\ &=\underbrace{gxg^{-1}} \underbrace{gyg^{-1}}\underbrace{gx^{-1}g^{-1}}\underbrace{gy^{-1}g^{-1}} \\ &=(gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1} \end{align}$$ is an element of $C$, so $C$ is normal.