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I'm trying to solve a textbook exercise asking me to show that $G =\langle x, y \space | \space x^2=e=y^2 \rangle$ (where $e$ is the identity) is infinite.

I assume a natural strategy would be to try to show that $xy$, $xyxy$, $xyxyxy$ and so on are all distinct elements of $G$.

If we assume $(xy)^n$, $(xy)^m$ are equal, then we would get that $(xy)^{n-m} = e$, but I'm not sure how to go about showing that this holds only when $n=m$, and hence that every $(xy)^n$ is distinct.

I'd appreciate any help or hints you could offer.

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marked as duplicate by Dietrich Burde group-theory May 6 at 11:35

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  • $\begingroup$ I believe there need to be more assumptions about the group $G$. For example if you know that $G$ is abelian then what you are trying to prove is simply false. $\endgroup$ – Mark May 6 at 10:28
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    $\begingroup$ @Mark You don't need more assumptions. The group $\langle x,y\mid x^2=e=y^2\rangle $ is perfectly well defined (and not abelian). $\endgroup$ – Arnaud D. May 6 at 10:31
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    $\begingroup$ @Mark This is a presentation of the group. It tells you everything about it. In particular you know it isn't abelian. $\endgroup$ – Matt Samuel May 6 at 10:31
  • $\begingroup$ Oh, I guess it is the group of all "formal words" in $x$ and $y$. I got it. $\endgroup$ – Mark May 6 at 10:32
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If you're allowed to use the dihedral group, you can say that if we add the relation $(xy)^m=1$ the resulting group has $2m$ elements, thus the group must be infinite since there are surjective homomorphisms onto all these groups.

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Maybe do it inductively. First of all, $xy$ is not $x, y$ or $e$. This can be seen by assuming an equality and then multiplying be $x$ or $y$, and using the fact that $y \ne x$.

Now, assuming that the set $\{x, y, (xy)^k |0<k<n\}$ has size $n+1$, we want to show this for $k \le n$. Assume for contradiction that there is some $m < n$ such that $(xy)^m = (xy)^n$, we get that

$(xy)^{n-m} = e$. But if $n \ne m$, then $0<n-m<n$, so $(xy)^{n-m}=e$ is a contradiction.

Edit: I believe this argument fails as noted in comments, since it does not prove that $x \ne y$.

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    $\begingroup$ But how do you know that $x \neq y$? $\endgroup$ – lisyarus May 6 at 10:42
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    $\begingroup$ Good point. Assuming $x = y$, we get that $x = y = e$, by multiplication by either $x$ or $y$, and therefore the group is trivial. This is a contradiction. For example, the smaller group where you remove $y$ as a generator is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, which is non-trivial. $\endgroup$ – Richard Jensen May 6 at 10:47
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    $\begingroup$ $x=y$ would not imply that the group is trivial, it would only imply that it is cyclic. $\endgroup$ – Arnaud D. May 6 at 10:50
  • $\begingroup$ Damn, I think you're right. I can't find a simple argument here, should there be one? $\endgroup$ – Richard Jensen May 6 at 11:02

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