How to show $G_{m}\cong G_n $ if and only if $n=m$, where $G_m:= \langle x,y \mid x(yx)^{m}=y(xy)^{m}\rangle$

Question

I have this family of groups $G_{m}:= \langle x,y \mid x(yx)^{m}=y(xy)^{m}\rangle$. I want to show that for different $m$ these two groups are either isomorphic or not. My guess is they are not and I tried to show that via abelization but of course you get a trivial group for all $m$'s so that doesnt work. Any ideas how to show this?

Answer 1

Good question. Every such group $G_m$ is an Artin group of spherical type with the corresponding diagram $D$ given by two vertices connected by an edge labeled $2m+1$.

$$ \begin{aligned} \circ\!\overset{2m+1}{--------}\!\circ \end{aligned} $$

(You can read in this Wikipedia article about Artin, or Artin-Tits, groups.) An Artin group is said to be of spherical type if the associated Coxeter group is finite. In your case, the associated Coxeter group of $G_m$ is a finite dihedral group: It is obtained by adding the relators $x^2=1, y^2=1$ to the presentation.

It is known that Artin groups of spherical type are isomorphic if and only if their diagrams are isomorphic, this is the main result in:

Paris, Luis, Artin groups of spherical type up to isomorphism, J. Algebra 281, No. 2, 666-678 (2004). ZBL1080.20033.

Hence, $G_n\cong G_m$ if and only if $n=m$.

Answer 2

Here is a computational solution to complement Mishe Kohan's geometrical solution.

I will use the rewritten presentation $G_m = \langle x,z \mid xz^mz=z^{m+1} \rangle$.

Since the abelianization of $G_m$ is ${\mathbb Z}$ for all $m$, it has a unique subgroup $H_m$ of index 2, generated by $a:=z$, $b:=x^2$, and $c:=xzx^{-1}$.

It is straightforward to use the Reidemeister-Schreier method to compute a presentation of $H_m$, and we get $$H_m \cong \langle \,b,c \mid c^mb=a^{m+1},ba^m=c^{m+1} \rangle,$$ which simplifies (eliminating $b$) to $$H_m \cong \langle a,c \mid a^{2m+1} = c^{2m+1} \rangle.$$ So $H_m$ has abelianization ${\mathbb Z} \oplus \frac{{\mathbb Z}}{(2m+1){\mathbb Z}}$.

So the groups $H_m$ and hence also $G_m$ are pairwise non-isomorphic for $m \ge 0$.

It is easy to see that $G_{-m} \cong G_{m-1}$ for $m \ge 0$, so negative $m$ do not give new groups.