Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
5  
That's all there is to it :-) –  Alex B. Nov 30 '11 at 15:57
2  
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. –  Thomas Andrews Nov 30 '11 at 15:57
    
thank you very much! –  user50948 Nov 30 '11 at 15:58
2  
You should also check that the quotient is actually a group (i.e., that $C$ is normal) if it hasn't been done already. –  user83827 Nov 30 '11 at 16:21
3  
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. –  Srivatsan Nov 30 '11 at 22:39

3 Answers 3

For some reason the OP won't post, or can't post, an answer, so summarizing the comments:

$(1)\,\,\forall\,g,x,y\in G\,$ , and putting $\,a^b:=b^{-1}ab\,\,,\,a,b\in G\,$: $$[x,y]^g:=g^{-1}[x,y]g:=g^{-1}x^{-1}y^{-1}xy g=\left(x^{-1}\right)^g\left(y^{-1}\right)^gx^gy^g=[x^g,y^g]\in G'\Longrightarrow G'\triangleleft G$$and thus the quotient $\,G/G'\,$ is a group.

$(2)\,\,$ Let now $\,N\,$ be any normal subgroup of $\,G\,$ s.t. $\,G/N\,$ is abelian, then: $$\forall\,x,y\in G\,\,,xNyN=yNxN\Longleftrightarrow xyN=yxN\Longrightarrow [x,y]:=(yx)^{-1}xy\in N$$ and since $\,G':=\langle\,[x,y]\;:\;x,y\in G\,\rangle\,$ , then $\,G'\leq N\,\Longrightarrow \,G'$ is the minimal (normal) subgroup of

$\,G\,$ s.t. its quotient is abelian -- "minimal" wrt set inclusion --.

Exercise: Explain the parentheses around "normal" above, i.e. show that any subgroup of G containing the commutator subgroup is normal.

share|improve this answer

Recall that $gh = hg[g,h]$ and that $[G,G]$ is the subgroup of $G$ generated by all commutators. $[G,G] \unlhd G$ because $$[h,k]g = g[h,k][[h,k],g].$$

The map $$G \longrightarrow \frac{G}{[G,G]}$$ is called abelianization (precisely because of the theorem we are about to prove).

Every element of $G/[G,G]$ is of the form $g [G,G]$ and this group is abelian because $$(g [G,G]) (g' [G,G]) = g g' [G,G] = g g' [g',g] [G,G] = g' g [G,G] = (g' [G,G]) (g [G,G]).$$

share|improve this answer

It is easy to show that the commutator subgroup is a characteristic subgroup , hence it is a normal subgroup. Proving G/C abelian is straightforward recalling that C is the subgroup of G generated by all commutators.

share|improve this answer
2  
This doesn't have much to do with the question. –  Tara B Feb 27 '13 at 20:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.