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Trying to get my head around the commutator subgroup. This is an excercise from Artin's Algebra:

Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian.

Here is what I've done:

Let $xC,yC \in G/C$ then $xyx^{-1}y^{-1}C = C$ since $xyx^{-1}y^{-1}$ is a commutator hence belongs to $C$. But then $xyC = yxC$ so $xC$ and $yC$ commute in $G/C$. This can be done for any elements, so $G/C$ is abelian.

This seems somewhat surprisingly short. Is that all there is to it?

Regards

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    That's all there is to it :-)2011-11-30
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    Yes. Basically, if $H$ is a subgroup of $G$ that contains $xyx^{-1}y^{-1}$ for some $x,y\in G$, then $xH$ and $yH$ commute in $G/H$. The commutator subgroup is just defined as the minimal subgroup that contains all elements of $G$ of the form $xyx^{-1}y^{-1}$, and hence it is the smallest subgroup such that $G/C$ is commutative.2011-11-30
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    You should also check that the quotient is actually a group (i.e., that $C$ is normal) if it hasn't been done already.2011-11-30
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    user19447, can you post a brief answer summarizing the comments and accept it? Thanks!2011-11-30
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    @Srivatsan well, I tried but it seems I don't have enough reputation to do so. Besides isn't it a bit redundant to post an answer?2011-11-30
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    I am not sure why you are not allowed to post an answer. To answer your question: On the contrary, it is good practise in this site to post an answer. This has several benefits, including: (1.) You'll have practise in writing the proof. Moreover others can check your work and upvote/comment on/criticize your answer. (2.) Future visitors need not dig through several comments to understand the answer. (3.) This question will show up as answered, which means one less unanswered question for the site.2011-11-30
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    @ThomasAndrews Here $H$ should be normal in $G$.2017-05-17

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