0
$\begingroup$

Let $G$ be a group and $G'=\langle [x,y] : x , y \in G\rangle $, where $[x,y]=x^{-1}y^{-1}xy$. I am trying to prove that $G/G'$ is an abelian group.

What I've done: $$X\in G/G' \Rightarrow X=xG' ; x\in G$$ $$Y\in G/G' \Rightarrow Y=yG' ; y\in G$$ $$XY=(xG')(yG')=xyG'$$ $$YX=(yG')(xG')=yxG'$$

Now, I have to prove that $XY=YX$, but I don't know how to do that :( Any help would be appreciated.

$\endgroup$
  • $\begingroup$ I know that G' is a normal subgroup of G, but I don't know that it can be useful or not... $\endgroup$ – User-123 May 29 '17 at 14:02
  • 1
    $\begingroup$ If $G'$ wer not a normal subgroup, then the quotient $G/G'$ would not be a group and your question would be meaningless. $\endgroup$ – Mariano Suárez-Álvarez May 29 '17 at 14:03
1
$\begingroup$

Instead of proving that $XY=YX$, prove that $XYX^{-1}Y^{-1}=1_{G/G'}$, to do so notice that: $$XYX^{-1}Y^{-1}=[x,y]G'.$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

To prove that $G/G'$ is abelian you need to show that $\forall x,y \in G$

$$(xG')(yG')=(yG')(xG') \iff xyG' = yxG' \iff x^{-1}y^{-1}xy \in G'$$

What is the definition of $G'$?

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Hint

$$yxG'=yx(x^{-1}y^{-1}xy)G'=xyG'$$

| cite | improve this answer | |
$\endgroup$

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