# Prove or disprove that $G = \langle{x,y\;|\; x^3, y^3, x^{13}, [x,y]=1}\rangle$ is trivial.

Prove or disprove that $$G = \langle x,y \mid [x,y]=x^3=y^3=x^{13}=1 \rangle$$ is trivial.

So the fact that $$x^3=x^{13}$$ means that the order of $$x$$ divides both $$3$$ and $$13$$, thus $$|x|=1$$ and so x is the identity element. So then this group presentation can be written as:

$$G = \langle y \mid [1,y]=y^3=1 \rangle$$ which means that $$y^3 = yy^{-1} = 1$$

So this group is either $$\Bbb Z_3$$ or trivial. But we also have that $$y^4=y$$, and therefore this group must be $$\Bbb Z_3$$. Is this correct?

edit: No, I have not yet proven anything. This group could still be trivial and satisfy $$y^4=y$$. Can somebody help me out here? Thanks!

• To show that $G$ is nontrivial, give a nontrivial map $G \to \mathbb{Z}_3$. – anomaly Apr 18 at 23:42

You're right in that the coprimality of $$3$$ and $$13$$ implies that $$|x|=1$$. Then we have $$\langle y\mid [1, y]=y^3=1\rangle,$$ like you say, but $$[1,y]=1^{-1}y^{-1}1y=y^{-1}y=1$$, so what we're left with is $$\langle y\mid y^3\rangle\cong \Bbb Z_3.$$