4
$\begingroup$

I'm trying to solve a problem in my textbook which asks me to identify the groups $G_1 = \langle x,y \space | \space x^3y=y^2x^2=x^2y\rangle$ and $G_2 = \langle x,y \space | \space xy=yx, x^4=y^2 \rangle$ from the given presentations.

For $G_1$, I'm pretty sure I can say that $x^3y=x^2y \implies x^3y(x^2y)^{-1} = e \implies x = e$ (where $e$ is the identity), and then $x^3y=y^2x^2 \implies y = y^2$ as $x=e$ to get $G_1 = \langle x,y \space | \space x=y=e \rangle \cong \{e\}$ (though I'd appreciate it if you could tell me if I have this wrong).

What I'm struggling with is trying to do the same sort of thing for $G_2$ - I can't see any way of getting this into a form where I can see the represented group.

I'd appreciate any help you could offer.

$\endgroup$

marked as duplicate by user1729 group-theory May 7 at 10:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $xy=yx$ in $G_2$, so it's a (finitely-generated) abelian group. What do these look like? $\endgroup$ – Jane Doé May 5 at 21:38
  • 4
    $\begingroup$ For $G_1$, you are correct; $x^3y=x^2y$ means $x=e$, so you get $y=y^2$, hence $y=e$. For $G_2$, note that $xy=yx$ means that the group is abelian, so every element can be written as $x^ay^b$ with $a,b\in\mathbb{Z}$. Now use the other relation to further simplify. $\endgroup$ – Arturo Magidin May 5 at 21:39
  • 2
    $\begingroup$ $G_2$ is also in this question $\endgroup$ – Arnaud D. May 7 at 9:04
  • $\begingroup$ The question about $G_2$ can be viewed as "find the abelianisation of the group $\langle x, y\mid x^4=y^2\rangle$", and so the techniques in the answers to this question are applicable here. $\endgroup$ – user1729 May 7 at 10:13
  • $\begingroup$ In the mean time, unless anyone has serious objections I'll close this question as a duplicate of @ArnaudD.'s link: the answer to the $G_2$ question can be found there, while the $G_1$ solution was covered (correctly) in the question. $\endgroup$ – user1729 May 7 at 10:16
1
$\begingroup$

$$\begin{align}G_2 &= \langle x, y | xy = yx, x^4 = y^2 \rangle \\ &= \frac{ \langle x, y | xy = yx \rangle }{\langle \langle x^4y^{-2} \rangle \rangle} \\ &= \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^4y^{-2} \rangle} \\ &\cong \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^2y^{-1} \rangle} \times C_2 \\ &\cong C_{\infty} \times C_2.\end{align}$$

$\endgroup$

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