# The Commutator Subgroup is Normal. [duplicate]

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 $C$.

## marked as duplicate by Dietrich Burde, Derek Holt abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 13 '17 at 21:15
• 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. – Claudius Mar 13 '17 at 14:21
• For the first bit, $gxyx^{-1}y^{-1}g^{-1}=gxygg^{-1}x^{-1}y^{-1}g^{-1}$. That might help. – Shaun Mar 13 '17 at 14:32
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.