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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.

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  • $\begingroup$ It's very easy if you know some elementary algebraic topology, but, I guess, you do not. $\endgroup$ – Moishe Kohan Oct 18 '13 at 17:45
  • $\begingroup$ @studiosus I'm aware of the covering space thing, please give a non-algebraic topology proof. Thank you very much. $\endgroup$ – HLC Oct 18 '13 at 19:57
  • $\begingroup$ What do you mean "no relations between them"? After all, $\;x^2=y^{-2}(y^2x^2)\;$...I think you must use in some way freeness, trivial relations, normal form, etc. $\endgroup$ – DonAntonio Oct 18 '13 at 22:20
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    $\begingroup$ I mean proving that the subgroup generated by $x^2$,$y^2$,$xy$ is free of rank 3. Or, if you define $\phi:<u,v,w>\rightarrow <x^2,y^2,xy>$ sending $u$ to $x^2$, $v$ to $xy$, $w$ to $y^2$, then the question is to show $\phi$ has trivial kernel. $\endgroup$ – HLC Oct 18 '13 at 22:50
  • $\begingroup$ HLC: Could you explain why do you need an algebraic proof if you already know a perfectly fine and intuitive topological argument? I can write an algebraic proof, but it will be ugly, like Schreier's proof that subgroups of free groups are free. (Reidemeister-Schreier algorithm.) $\endgroup$ – Moishe Kohan Oct 19 '13 at 12:45
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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.

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