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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?

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  • $\begingroup$ "Unequal" is not what you mean. The subgroups $\langle (12)\rangle$ and $\langle (23)\rangle$ of the symmetric group $S_3$ are isomorphic but not equal. $\endgroup$ – YCor Jun 13 '19 at 14:45
  • $\begingroup$ Well, by "unequal" I meant that sending generators on generators ($a\mapsto a'$, $b\mapsto b'$) does not extend to an isomorphism. Anyway, that's why I added the subsequent clarifying sentence. $\endgroup$ – Daniel Jun 13 '19 at 14:52
  • $\begingroup$ OK, it's sometimes called "isomorphic as marked groups". $\endgroup$ – YCor Jun 13 '19 at 14:58
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    $\begingroup$ As it seems a reasonable guess, I would try to find a representation of the n- group without the m condition. There are at least two ways to think about this: 1. Find a subgroup of S_n that makes the work. 2. Find a subgroup of matrices that makes the work. Without compitations, you can think about a group action (linear or not) where m-condition is not satisfied. $\endgroup$ – Andrea Marino Jun 13 '19 at 23:33
  • $\begingroup$ A more combinatorial approach. Take the free group on 3 generators $F_3$ and the canonical map to $G(l,k,n)$. Let $T$ be the kernel. It is generated by $a^l, b^k, \sigma:=(ab)^n(a^{-1}b^{-1})^n$.We want $\tau:=(ab)^m(a^{-1}b^{-1})^m \not \in T$. Call $G=\langle a^l, b^k \rangle , H= \langle\sigma \rangle $. Evidently $\tau \not \in G$, so that possibly $\tau = g_1 h_1 \ldots h_n g_n$; it may have different endings, but there is one $h$. I then claim that $\tau$, when simplified, contains $b(ab)^{n-1}(a^{-1}b^{-1})^{n-1}a$, morally because high and low exponents alternate and cannot simplify. $\endgroup$ – Andrea Marino Jun 14 '19 at 0:07
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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 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 $n$.

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  • $\begingroup$ About your second to last sentence: the rotation angles of $ab$ and $ba$ are the same. $\endgroup$ – Luc Guyot Jun 19 '19 at 16:45

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