This comes from Artin Second Edition, page 219. Artin defined $G=\langle x,y\mid x^3, y^3, yxyxy\rangle$ , and uses the Todd-Coxeter Algorithm to show that the subgroup $h=\langle y\rangle$ has index 1, and therefore he gets $G=H$ is the cyclic group of order 3.
My question is: why $\langle y\rangle$ is not trivial? The Todd-Coxeter Algorithm gives no extra information about $y$ except it acts trivially on $\langle y\rangle$ as any other elements in $G$ do. So how do we know that $y\neq 1$?
Many thanks!
Best Answer
The group $G$ is cyclic of order $3$, generated by the element $y$.
Here are two proofs. Both are pretty similar, the main difference is how you conclude that the group is non-trivial (implicit in the first, explicit in the second).
First proof: Presentations. Add in the new generator $z:=yx$, and then remove $x$ via the relevant Tietze transformation to get: $$\langle y, z\mid (y^{-1}z)^3, y^3, z^2y\rangle.$$ We then see that the group is cyclic, generated by $z$ (as $y=z^{-2}$). So lets find the relevant presentation: $$\langle z\mid (z^3)^3, (z^2)^3\rangle=\langle z\mid z^9, z^6\rangle.$$ This simplifies to: $$\langle z\mid z^3\rangle.$$ Then it is generated by $y$ as $z=yx$ is a generator, and therefore so is $y=(xy)^{-2}$.
Second proof: Use the abelianisation. By the relation $yxyxy$, we see that $y^{-1}\in\langle yx\rangle$, and so $x, y\in\langle yx\rangle$. Therefore, $yx$ generates our group $G=\langle x, y\rangle$. This means that $G$ is abelian, so $(yx)^3=y^3x^3=1$. Therefore, $G$ is either cyclic of order three or trivial. As the comments to the question point out, the map which kills $x$ has non-trivial image, and so $G$ is non-trivial. Therefore, $G$ is cyclic of order $3$, and generated by $yx$. Conclude that $G=\langle y\rangle$ as in the first proof (or by noting that the map which kills $x$ much be an isomorphism, so $x$ is trivial and so $yx=y$).