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This question already has an answer here:

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 :)

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marked as duplicate by Dietrich Burde, Derek Holt abstract-algebra Mar 13 '17 at 21:15

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    $\begingroup$ If $x$ lies in the commutator and $g\in G$, then $gxg^{-1} = (gxg^{-1}x^{-1})\cdot x$ lies in the commutator. Hence the commutator subgroup is normal. $\endgroup$ – Claudius Mar 13 '17 at 14:21
  • $\begingroup$ For the first bit, $gxyx^{-1}y^{-1}g^{-1}=gxygg^{-1}x^{-1}y^{-1}g^{-1}$. That might help. $\endgroup$ – Shaun Mar 13 '17 at 14:32
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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.

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