# Group with a given presentation, finite or infinite?

Consider the group with following presentation,

$$G=\langle s,t : s^2=1, (st)^{3}=1\rangle$$

Is this group finite or infinite?

I tried to manipulate the relations and could only get $$(ts)^3=1$$. I don't know how to proceed further. Any hints?

• Doesn't $t$ have infinite order in this group? – Greg Martin Dec 3 at 7:06
• @GregMartin I thought the same thing. But I cannot prove it. – Abhikumbale Dec 3 at 7:07

Hint: Instead of taking $$s$$ and $$t$$ as generators, take $$s$$ and $$st$$ as generators. How else can you describe the group then?
Writing $$u=st$$, we have $$G=\langle s,u\mid s^2=1, u^3=1\rangle$$. But this just means that $$G$$ is the free product of a cyclic group of order $$2$$ (generated by $$s$$) and a cyclic group of order $$3$$ (generated by $$u$$). In particular, $$G$$ is infinite, because for instance there are infinitely many distinct reduced words of the form $$sususu\dots$$.
• Its not immediately obvious that free products are infinite (I mean, you have to prove that no two reduced sequences are equal). One way to do this is to let the group act on the infinite tree where every vertex has degree two or three, and no two vertices of the same degree are incident. The action is: fix a vertex $V_s$ of valency $2$ and another $V_u$ of valency $3$ then $s$ acts by rotating the tree around $V_s$ while $u$ acts by rotating the tree around $V_u$. Then find the paths in the tree corresponding to these products, and note that they are non-equal for non-equal products. – user1729 Dec 3 at 11:00
• @user1729 Take two rotations of the plane, of order $k\ge 2$ and $\ell\ge 2$. They have no common fixed point, hence generate an infinite group. So the free product of any two nontrivial cyclic groups is infinite. – YCor Dec 3 at 13:32