# Proving some finitely presented groups being unequal

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.

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?

• "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. – YCor Jun 13 '19 at 14:45
• 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. – Daniel Jun 13 '19 at 14:52
• OK, it's sometimes called "isomorphic as marked groups". – YCor Jun 13 '19 at 14:58
• 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. – Andrea Marino Jun 13 '19 at 23:33
• 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. – Andrea Marino Jun 14 '19 at 0:07

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 $$n$$.

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