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Let's look at the following presentation: $$ \Delta^*(p,q,r;s/2)=\langle a,b,c\mid a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=((abc)^2)^{s/2}=1\rangle $$ This is a presentation of a special triangle group $\Delta^*(p,q,r; s/2)$.

Focusing on the corresponding index-two subgroup of $\Delta^*$ (Von Dyck group), we get $$ \Delta_0^*(p,q,r; s/2)=\langle x,y,z\mid x^p=y^q=z^r=xyz=(xzy)^{s/2}=1\rangle , $$ where $x=ab, y=bc, z=ca$ (we see $xyz=ab\;bc\;ca=1$) and $xzy=ab\; ca\; bc=(abc)^2$. It is said, that this means that $\Delta^*_0(p,q,r;s/2)$ is a discrete group consisting of orientation-preserving isometries of the hyperbolic plane. I think this relates e.g. to the regular triangles-tilings of the hyberbolic plane (correct me if I'm wrong)...

Does $(xzy)^{s/2}$ preserve some special kind of property like orientation?

The presentation in question are motivated by this and that...

draks ...
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    If you cancel the $z$-generator by replacing it with $y^{-1}x^{-1}$ via Tietze transformations in the usual way, then $xzy\mapsto xy^{-1}x^{-1}y$. So we see that your group has presentation $$\langle x, y\mid x^p, y^q, (xy)^r, [x,y]^{s/2}\rangle.$$ In particular, if $s=2$ then your group is finite abelian, while if $p, q, r, s>6$ then (I think) your group is $C'(1/6)$ small cancellation (so infinite and hyperbolic, etc.). – user1729 Sep 09 '20 at 09:48
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    What triangle group is this? not seeing it in the page you link to. Is it actually a quotient of a triangle group and you are wondering what that element did in the hyperbolic plane before quotienting(when the numbers correspond to a hyperbolic triangle group)? –  Sep 09 '20 at 10:43
  • @Paul 1. maybe Derek's answer helps; 2a. ok; 2b. I thought $xyz=1$ makes up the orientation preservation of triangles shuffled/permuted around in the hyperbolic plane, by the actions of $x,y,z$ (is that correct?). So I wonder what $(xzy)^{s/2}$ can tell us about the triangle shuffling/permutation in terms of properties like e.g. orientation preservation... – draks ... Sep 09 '20 at 10:47
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    Well, product of orientation preserve orientation. Saying $xyz=1$ says that $xyz$ does nothing. If your group isn't a hyperbolic triangle group then the group probably isn't acting on the plane in any natural way. –  Sep 09 '20 at 11:23
  • @PaulPlummer ok I see. 1. So you say that the additional relation $(xzy)^{s/2}=1$ make my group a quotient group, right? Would it be quotient groups of "ordinary" triangle groups as stated here? 2. And because of this "quotient of a triangle group" thing it does not act on the hyperbolic plane in a natural way. Can you explain why's that? – draks ... Sep 09 '20 at 14:50
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    @draks Actions are about homomorphisms: An action of $G$ on $X$ corresponds to a homomorphism $\alpha: G\rightarrow\operatorname{Aut}(X)$. So, for example, if we have a homomorphism $\beta: H\rightarrow G$ then we have an action of $H$ on $X$ defined by $\alpha\beta:H\rightarrow\operatorname{Aut}(X)$. However, in your setting you have $G\rightarrow\operatorname{Aut}(X)$ and $G\rightarrow H$ (where $G=\Delta_o$ and $H=\Delta_o^$). So there is no reason* that there should exist an action $H\rightarrow\operatorname{Aut}(X)$. So we wouldn't expect an action of $\Delta_o^*$ on your space. – user1729 Sep 09 '20 at 16:13
  • @user1729 ok thanks, but as far as I understood your answer here, isn't $H=G/N$ a quotient group with a normal subgroup $N$? If so, wouldn't that imply an action on $X$ via the cosets? – draks ... Sep 11 '20 at 05:55
  • @draks... My answer there is the same situation as my comment above. So no, there is no reason that this set up would imply an action on $X$ via the cosets. (If you want further clarification then I suggest you ask a new question, about the abstract set up, without mentioning triangle groups. But really, it is covered in the last paragraph of Paul's answer, including counter-examples.) – user1729 Sep 11 '20 at 08:58

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As far as I can tell your $\Delta^*(p,q,r;s/2)$ isn't a triangle group, it is a quotient of the triangle group $\Delta(p,q,r)$ by adding the relation $((abc)^2)^{s/2}=1$ (It could secretly be a triangle group but I doubt it). You could say it is some "generalized triangle group" if you want but then the connection to acting on hyperbolic plane etc. starts to break.

For now just look $\Delta_0(p,q,r)$ von Dyck group with respect to the same generators you give. Every element in $\Delta_0$ preserves orientation, almost by definition, which includes $xzy$ and any power of that element. Now remember the group you describe is a quotient of the above group and you set $(xzy)^{s/2}=1$, so that element does nothing to any space you act on. If you want you can think of as orientation preserving as it "preserves" everything

Generally, if a group $G$ acts on a space $X$ then quotient groups will not, necessarily, have a natural action on $X$. As a simple example, $\mathbb{Z}$ acts on $\mathbb{R}$ by translation or you can have $\mathbb{Z}$ act on the hyperbolic plane by choosing a hyperbolic or parabolic isometry. Any action of $\mathbb{Z/7Z}$, a quotient group of $\mathbb{Z}$, on $\mathbb{R}$ must fix a point and in this case there is no nontrivial action. Similarly $\mathbb{Z/7Z}$ would have to fix a point on the hyperbolic plane, so it could be a rotation of order seven but a rotation doesn't really have anything to do with original action. Note that $\mathbb{Z}$ in the above cases does not fix any point in $\mathbb R$ or the hyperbolic plane. With that being said I don't think $\Delta_0^*$ generally has (natural) actions coming from the group $\Delta_0$.

draks ...
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    They are not, in general, triangle groups. For example, take $p, q$ each to be odd, then $\Delta_o^*(p, q, pq; 1)$ is simply $\mathbb{Z}_p\times \mathbb{Z}_q$, which has odd order. However, finite triangle groups all have even order (as if $1/p+1/q+1/r>1$ then one of $p,q,r$ must be $2$, so a generator has even order). – user1729 Sep 09 '20 at 16:37
  • +1 thanks so far, I'm starting to understand, but: "Any action of $\mathbb{Z/7Z}$ on $\mathbb{R}$ must fix a point and in this case there is no nontrivial action." 1. why must it fix a point? Is this a kind of identity element? Can a point be a subset or a coset of the group (I don't feel well with that question, hope you don't mind)? 2. how do nontrivial actions look like? – draks ... Sep 11 '20 at 14:54
  • @draks... I was thinking about acting on $\mathbb{R}$ by isometires, so in order to act you must send $1 \in \mathbb{Z}/7\mathbb{Z}$ to some isometry. That is you need a homomorphism $ \mathbb{Z}/7\mathbb{Z} \to \mathrm{Iso}(\mathbb{R})$. Isometries of $\mathbb{R}$ are just reflections or translations. Translations can't have finite order, reflections have order two and fix a point. Group theory exercise: you can't map $1 \in \mathbb{Z}/7\mathbb{Z}$ to an order two element. The same actually is true if you just act by homeomorphisms (and not much harder if you know a some of topology). –  Sep 11 '20 at 17:12
  • Nontrivial actions are actions with at least one element which acts nontrivially or you can think of the image of the homemorphism $G \to \mathrm{Aut}(X)$ is more than just the identity. In the case of actions on $\mathbb{R}$ by isometries, to be nontrivial at least one element needs to move some point. –  Sep 11 '20 at 17:17
  • Do you mean the $\mathbb Z / 7\mathbb Z$ or $\mathbb Z / 7\mathbb Z^\times$? Your nomenclature and "Translation in $\mathbb R$" rather feels like the additive one, but this makes $1\in \mathbb Z / 7\mathbb Z$ look like a peculiar choice... and I bet that homemorphism is homomorphism, right? Just because you also spoke about homeomorphisms before... – draks ... Sep 11 '20 at 18:31
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    @draks... It is additive. A homomorphisms from a cyclic group/action of cyclic group is determined by what a generator does (1 is a nice choice of generator). The homeomorphism is meant to be homeomorphism not "homomorphism". It was just a a side remark that $\mathbb{Z}/7\mathbb{Z}$ must act trivially if the action is by isometries or, more generally, by homeomorphisms. I only really talked about isometries, but the homeomorphism case isn't much harder. –  Sep 11 '20 at 18:45
  • ahhh... I gotta think about that... – draks ... Sep 11 '20 at 18:51