How to show that elements in these groups pairwise commute?

Question

Let $F_2$ be the free group on $x,y$ and $N$ the normal subgroup generated by $x^2,y^4,xyx^{-1}y^{-1}$. Consider the group $F_2/N$ and two elements $xN$ and $yN$ in it. How do I show that elements in the subgroups generated by $xN$ and $yN$ pairwise commute?

Answer 1

Recall, that you've got a well defined group-homomorphism $x\mapsto xN$ (namely, the quotient map), which implies $(xN)(yN)=(xy)N$. For simiplicity, I'll avoid using the notation $xN$ for the elements of the quotient, and I'll just write $x$.

Note that in $F_2/N$, you have $xyx^{-1}y^{-1}=1$, so $xy=yx$ in $F_2/N$ for all $x,y\in F_2$. Since $x^2=1$ and $y^4=1$ in $F_2/N$ you only have to check that $x$ commutes with $y^2$ and $y^3$, which is immediate since $xy^2=xyy=yxy=y^2x$, and the same works for $y^3$.