4
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ "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
    Commented Jun 13, 2019 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
    Commented Jun 13, 2019 at 14:52
  • $\begingroup$ OK, it's sometimes called "isomorphic as marked groups". $\endgroup$
    – YCor
    Commented Jun 13, 2019 at 14:58
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jun 13, 2019 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. $\endgroup$ Commented Jun 14, 2019 at 0:07

1 Answer 1

2
$\begingroup$

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 $n$.

$\endgroup$
1
  • $\begingroup$ About your second to last sentence: the rotation angles of $ab$ and $ba$ are the same. $\endgroup$
    – Luc Guyot
    Commented Jun 19, 2019 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.