# Show that $G =\langle x, y \space | \space x^2=e=y^2 \rangle$ is infinite [duplicate]

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.

## marked as duplicate by Dietrich Burde group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 6 at 11:35

• 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. – Mark May 6 at 10:28
• @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). – Arnaud D. May 6 at 10:31
• @Mark This is a presentation of the group. It tells you everything about it. In particular you know it isn't abelian. – Matt Samuel May 6 at 10:31
• Oh, I guess it is the group of all "formal words" in $x$ and $y$. I got it. – Mark May 6 at 10:32

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.

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 $$\{x, y, (xy)^k |0 has size $$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 $$0, so $$(xy)^{n-m}=e$$ is a contradiction.

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

• But how do you know that $x \neq y$? – lisyarus May 6 at 10:42
• 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. – Richard Jensen May 6 at 10:47
• $x=y$ would not imply that the group is trivial, it would only imply that it is cyclic. – Arnaud D. May 6 at 10:50
• Damn, I think you're right. I can't find a simple argument here, should there be one? – Richard Jensen May 6 at 11:02