TITLE: Prove elements generate a free group
QUESTION [17 upvotes]: How does one show that the elements $x^2$, $y^2$, and $xy$ have no nontrivial relations among them in the free group generated by $\{x,y\}$?  This would prove that the free group $F_2$ has a subgroup isomorphic to $F_3$.
(I understand the problem, it's analogous to proving the linear independence of three vectors... except in this case the non-commutativity of the basis is the crucial piece, rather than just the multiplicity of the basis elements.  I just need a hint to get me in the right direction...)
Source is Artin's "Algebra" 1ed #6.7.2, not homework just learning.

REPLY [3 votes]: Instead of the choice in the OP, we could select differently in order to get wisdom of the problem with respect to "orderings of words".
Among the 12 possibilities of length two reduced words in $F_2$ we can prove
that the the first three in the ordering:
$$x^2<xy<xy^{-1}<x^{-2}<x^{-1}y<x^{-1}y^{-1}<yx<yx^{-1}<y^2<y^{-1}x<y^{-1}x^{-1}<y^{-2}$$
generate the other nine.
Let $S=\{ x^2\ ,\ xy\ ,\ xy^{-1}\}$  be ours alternative choice, then we get $S^{-1}=\{ x^{-2}\ ,\ y^{-1}x^{-1}\ ,\ yx^{-1}\}$.
Now, there are six remaining cases:

$x^{-1}y=x^{-2}(yx^{-1})^{-1}y^2$  then $y^{-1}x=y^{-2}(yx^{-1})x^2$
$x^{-1}y^{-1}=x^{-2}(xy)y^{-2}$ then $yx=y^2(xy)^{-1}x^2$
$y^2=(yx^{-1})(xy)$ then $y^{-2}=(xy)^{-1}(yx^{-1})^{-1}$,

that is, the product of the first three reduced words of length two and their inverses generate all the twelve reduced words of length two.
So, any other word of even length in $F_2$ can be generated by $S\cup S^{-1}$.
Now, is a matter of following the Schreier's algorithm to warrant that the subgroup $U=\langle\{ x^2\ ,\ xy\ ,\ xy^{-1}\}\rangle$ is indeed free. 
For, we must be sure that $F=U\cup Ux$ is a partition.
This is the case because if one admits that $U$ carries all the even length words, then $Ux$ carries the odd length words. Also the presence of $xy^{-1}\in U$ implies $Ux=Uy$.
So a set of transversals is $\Sigma=\{1,x\}$, which they are minimal in their coset and which together with  $S$ they are going to give us the set 
$$\Sigma S=\{\ x^2\ ,\ xy\ ,\ xy^{-1}\ ,\ x^3\ ,\ x^2y\ ,\ x^2y^{-1}\ \},$$ 
hence one receives $\overline{\Sigma S}=\{1,x\}$.
According to Schreier's method the set $\{gs\overline{gs}^{-1}\ |\ g\in\Sigma,s\in S\}$ carries the free generator for $U$. In our case one computes:
\begin{eqnarray*}
x^2\overline{x^2}^{-1}&=&x^2,\\
xy\overline{xy}^{-1}&=&xy,\\ 
xy^{-1}\overline{xy^{-1}}^{-1}&=&xy^{-1},\\ 
x^3\overline{x^3}^{-1}&=&x^2,\\
x^2y\overline{x^2y}^{-1}&=&x^2yx^{-1},\\ 
x^2y^{-1}\overline{x^2y^{-1}}^{-1}&=&x^2y^{-1}x^{-1}
\end{eqnarray*}
Which they all are in $U$.
Hence  $\{\ x^2\ ,\ xy\ ,\ xy^{-1}\ \}$ is a set of free generator for $U$.