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

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

• That's all there is to it :-) – Alex B. Nov 30 '11 at 15:57
• 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
• 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
• user19447, can you post a brief answer summarizing the comments and accept it? Thanks! – Srivatsan Nov 30 '11 at 19:02

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.

• How can I find the order of $G/G'$? – Salvattore May 31 '16 at 22:21

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.

• This doesn't have much to do with the question. – Tara B Feb 27 '13 at 20:48