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!

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

The tools used here are known as Tietze transformations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.