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.
Best Answer
This proof uses covering spaces, but I'm posting it at the request of orlandpm (the OP).
Consider the $2$-oriented graph, $G$:
This space is the wedge sum of two circles. Therefore its fundamental group is the free group on two generators $\langle x, y\rangle$.
This space has the following covering space, $\widetilde G$:
Since the map $p_* : \pi_1(\widetilde G) \to \pi_1(G)$ induced by the covering space $p : \widetilde G \to G$ is injective, it follows that the fundamental group of $\widetilde G$ is isomorphic to $\langle x^2, xy, y^2\rangle$.
At the same time, $\widetilde G$ is homotopy equivalent to the wedge sum of three circles. Therefore its fundamental group is the free group on three generators, $F_3$.
We conclude that $$ \langle x^2, xy, y^2\rangle \cong F_3. $$
For the theory behind this, check out section $1.3$ of Hatcher's Algebraic Topology book (available freely online). (Images courtesy of the book.)