free groups: no relations between generators Consider the free group $\langle x,y\rangle$. I'd like to show that $x^2,y^2,xy$ have no relations between them, without the theorem that a subgroup of a free group is free, without knowledge about rank (but any universal property is allowed to use of course). Is there a neat way? Thanks.
 A: Consider the free group $F_3$ of rank 3 with generators $a,b,c$. Consider the automorphism $\phi$ given by $a\mapsto a$, $b\mapsto cbc^{-1}$, $c\mapsto abc^{-1}$. Consider the corresponding semidirect product $G=\mathbf{Z}\ltimes F_3$, where the generator $t$ of $\mathbf{Z}$ acts by $tgt^{-1}=\phi(g)$ for $g\in F_3$. The group $G$ has thus an obvious presentation with the generating set $\{t,a,b,c\}$ and 3 relators $tgt^{-1}=\phi(g)$ for $g=a,b,c$.
Note that $\phi^2$ maps $a\mapsto a$, $b\mapsto aba^{-1}$, $c\mapsto aca^{-1}$ and thus is the inner automorphism by $a$. So the element $t^2a^{-1}$ is central in $G$. If we mod out by the (central) cyclic subgroup $Z$ generated by this element and replace $a$ by $t^2$ in the relators, the group $H$ we get has the presentation with generators $t,b,c$ and relators $tbt^{-1}=cbc^{-1}$ and $tct^{-1}=t^2bc^{-1}$. The first relator reads as $[t^{-1}c,b]=1$ and last relator reads as $b=(t^{-1}c)^2$. Thus the first relator is redundant, and the last relator just means that $b$ is a redundant generator and that $H$ is the free group on the two generators $t,c$. 
There is a homomorphism $\psi:G\to F_2$ mapping $t\mapsto x$, $a\mapsto x^2$, $b\mapsto y^2$, $c\mapsto xy$ (because they satisfy the 3 relators). Obviously this homomorphism is surjective, and factors through a homomorphism $\bar{\psi}:H\to F_2$ (since $t^2a^{-1}$ is in the kernel). The homomorphism $\bar{\psi}$ from $H$ to $F_2$ maps the basis $(t,b)$ to $(x,xy)$, which is a basis of $F_2$; hence $\bar{\psi}$ is an isomorphism. This means that $\psi$ has kernel $Z$. Since $Z\cap F_3$ is trivial, this implies that $\psi$ is injective in restriction to $F_3$. This is exactly what you wanted to prove.
