Prove that $\langle x,y,|x^2,y^2\rangle$ is an infinite group

Question

Here is my attempt: $xyx \neq xyxy \neq xyxyx \neq xyxyxy....$ gives infinite many elements in the group. Is this correct?

Answer 1

You're on the right track! Let $\text{FG}(x,y)=\langle x,y\rangle$, $G=\langle x,y\mid x^2,y^2\rangle$, and $H$ be the minimal normal subgroup of $\text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $\eta:\text{FG}(x,y)\to G$.

We know that every element of $\text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $g\in \text{FG}(x,y)$ denote this presentation as $\overline{g}$.

Consider the following algorithm $A$: given $g\in \text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $\overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxx\mapsto x\overbrace{yy}xx\mapsto \overbrace{xx}x\mapsto x$$ $$A:x^{-1}yyxxx\mapsto x\overbrace{yy}xxx\mapsto \overbrace{xx}xx\mapsto \overbrace{xx}\mapsto \text{empty}$$ Where "$\text{empty}$" is the empty string.

Denote $H'$ the subset of elements $\text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $\text{FG}(x,y)$ and in particular a normal subgroup of $\text{FG}(x,y)$. Let $\eta':\text{FG}(x,y)\to \text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $\eta'$ factors through a morphism $\kappa:G\to \text{FG}(x,y)/H'$ in the sense that $\eta'=\kappa\circ \eta$. Our $\kappa$ is therefore surjective (since $\eta'$ is surjective). So it suffices to demonstrate that $\text{FG}(x,y)/H'$ is infinite.

To that end consider the sequence of elements $$\eta'(xyx) \neq \eta(xyxy)' \neq \eta'(xyxyx) \neq \eta'(xyxyxy)\dots$$ We will prove that the elements of this sequence are pairwise distinct. Let $\eta'(g_1)$ and $\eta'(g_2)$ be two such elements. It suffices to show that $\eta'(g_1)\eta'(g_2)^{-1}\neq 1\iff A(g_1g_2^{-1})\neq \text{empty}$. Can you finish from here?

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It's worth noting that $H=H'$ and thus that $G\cong \text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $H\subseteq H'$, proving that $H'\subseteq H$ would do the trick.)