Let $F$ be the free group on $x$, $y$ and $R$ be the smallest normal subgroup containing the commutator $xyx^{-1}y^{-1}$.

a) Show that $x^2y^2x^{-2}y^{-2}$ is in $R$.

b) Prove that $R$ is the commutator subgroup of $F$.

I tried to conjugate $xyx^{-1}y^{-1}$ with some element in $F$ to get $x^2y^2x^{-2}y^{-2}$ but failed..


$R$ is the smallest "normal" subgroup containing $[x,y]$. Then in $G/R$ the images of $x$ and $y$ will commute, i.e. $G/R$ is abelian, i.e. $G'\subseteq R$. On the other hand, by definition of $R$, it is clear that $R\subseteq G'$.

a) can be solved with the help of b): in $G/R$ the element $x^2y^2x^{-2}y^{-2}$ goes to identity, hence it is in $G'=R$.

The other way: we write $[a,b]=aba^{-1}b^{-1}$. Then $$x^2y^2x^{-2}y^{-2}=x.(xy^2x^{-1}).x^{-1}y^2=x.(xy^2x^{-1}y^{-2}).y^2.x^{-1}y^2$$ $$=x.[x,y^2].y^2x^{-1}y^{-2}=x.[x,y^2].y^2x^{-1}y^{-2}x.x^{-1}=x.[x,y^2].[y^2,x^{-1}]x^{-1}$$ The last expression is conjugate of product of two commutators. It is sufficient to show that these two commutators are in $R$. You may try this easier computation than the original.

  • $\begingroup$ Hi Groups, thanks a lot for your answer. I am not clear about "R is the smallest "normal" subgroup containing [x,y]. Then in G/R the images of x and y will commute". Why commute? $\endgroup$ – Wei Ye Sep 28 '15 at 14:08
  • $\begingroup$ I found group $G=<x, y; xyx^{-1}y^{-1}>$ is called a free abelian group, that make $G/R$ abelian. $\endgroup$ – Wei Ye Oct 8 '15 at 2:29

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