Let me define the following groups
$$G(k,l,n):=\langle a,b\mid a^k=1=b^l, (ab)^n=(ba)^n\rangle$$


Fixed $k$ and $l$ (WLOG we can assume those are prime), I would like to know, whether the groups are mutually unequal. More precisely, prove that $(ab)^m\neq (ba)^m$ for $m<n$.


For $k=l=2$ those are the dihedral groups, so the answer is yes. For any other $k$ and $l$, I do not know.


I am not a group theorist, so I have no feeling for such a question, so I do not know, whether it is obvious, or it is probably true but hard to prove, or whether it is totally unclear. I would be happy even for any comment on this.


I was trying to find similar groups in the literature. I encountered the von Dyck groups defined as
$$D(k,l,n)=\langle a,b\mid a^k=b^l=(ab)^n=1\rangle.$$
(Again, for $k=l=2$, those are dihedral groups.) Is some similar result known for those von Dyck groups?



Proving some finitely presented groups being unequal
Ok, I think I have a proof of this. First of all, note that $D(k,l,n)$ is a quotient of $G(k,l,n)$ with respect to $(ab)^n=1$ (indeed, we have $(ba)^n=b(ab)^nb^{-1}=1=(ab)^n$ in $D(k,l,n)$). So, it is enough to prove in $D(k,l,n)$ that $(ab)^m\neq (ba)^m$. Now, we know that $D(k,l,n)$ [acts](https://en.wikipedia.org/wiki/Triangle_group) on Euclidean plane, the two-dimensional sphere, the real projective plane, or the hyperbolic plane. In particular, $ab$ generates a rotation by the angle $2\pi/n$ around some point, whereas $ba$ generates a rotation with respect to the opposite angle arround a *different point*. Hence $(ab)^m\neq (ba)^m$ unless $m$ is a multiple of $n$.


