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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!

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  • $\begingroup$ To show that $G$ is nontrivial, give a nontrivial map $G \to \mathbb{Z}_3$. $\endgroup$ – anomaly Apr 18 at 23:42
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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 $$\langle y\mid y^3\rangle\cong \Bbb Z_3.$$

The tools used here are known as Tietze transformations.

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