TITLE: 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 [8 upvotes]: 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?

REPLY [8 votes]: 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.