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?


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    $\begingroup$ That's all there is to it :-) $\endgroup$
    – Alex B.
    Commented Nov 30, 2011 at 15:57
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    $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2011 at 15:57
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    $\begingroup$ You should also check that the quotient is actually a group (i.e., that $C$ is normal) if it hasn't been done already. $\endgroup$
    – user83827
    Commented Nov 30, 2011 at 16:21
  • $\begingroup$ user19447, can you post a brief answer summarizing the comments and accept it? Thanks! $\endgroup$
    – Srivatsan
    Commented Nov 30, 2011 at 19:02
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    $\begingroup$ 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. $\endgroup$
    – Srivatsan
    Commented Nov 30, 2011 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: $$\forall\,x,y\in G,\,\,xNyN=yNxN\Longleftrightarrow xyN=yxN \Longleftrightarrow (yx)^{-1}xy\in N \Longleftrightarrow [x,y] \in N$$ 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.

  • $\begingroup$ How can I find the order of $G/G'$? $\endgroup$
    – Salvattore
    Commented May 31, 2016 at 22:21
  • $\begingroup$ something else came to my mind. There is a theory $G/N$ is abelian iff $G′$ is a subgroup of $N$ $\endgroup$
    – Napu
    Commented Apr 15, 2021 at 20:25

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 $$(g [G,G]) (g' [G,G]) = g g' [G,G] = g g' [g',g] [G,G] = g' g [G,G] = (g' [G,G]) (g [G,G]).$$


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.

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    $\begingroup$ This doesn't have much to do with the question. $\endgroup$
    – Tara B
    Commented Feb 27, 2013 at 20:48

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