Showing that a group acts freely and discretely on real plane

[kmitza]

TL;DR

I want to show that group acts freely and discreetly on IR^2 but I am failing to see what the group looks like and I can't describe the orbits

So before I start I technically do now that the group I am dealing with is just a representation of the Klein bottle but I am not supposed to use that as a fact because the goal of the problem is to derive that information.

Problem:
Let G be a group of with two generators a and b such that aba = b. Show that formulas $a(x,y) = (x,y+1)$ and $b(x,y) = (x+1,1-y)$ determine a free and discrete action of G on $\mathbb{R}^2$

So the problem I have is that I don't know what the relation of the generators gives me except that all strings of form $aba$ are reduced to "b" and similarly for the inverses. I can see that any part of an orbit obtained by just applying "a" or "b" is going to be discrete but I don't know how to translate it to an arbitrary member of the group. I had an idea of showing that the orbit is discrete because both actions of generators can be represented as translations over the integer lattice for $(x,y)$ and (x,-y) ## respectively, however this comes to bite me in the ass later as I need to show that there is no identification on the interior of the unit square so I really need to find a different way to calculate the orbit.

This is a HW problem so please don't give me full solutions immediately I am just in need of a hint or a nudge in a right direction

 

[anuttarasammyak]

I am sorry if I do not catch you in a right way, I observe
b^2(x,y)=(x+2,y)

 

[kmitza]

Um yes that is correct but it doesn't really help me, in general you can say that $b^n(x,y) = (x+n,y)$ if n is odd and $(x+n,1-y)$ if n is even but without knowing the structure of the group elements I don't know how to calculate the whole orbit. I think that all members should be of the form $a^nb^m$ but I am not completely sure about it

 

[kmitza]

Um yes that is correct but it doesn't really help me, in general you can say that $b^n(x,y) = (x+n,y)$ if n is odd and $(x+n,1-y)$ if n is even but without knowing the structure of the group elements I don't know how to calculate the whole orbit. I think that all members should be of the form $a^nb^m$ but I am not completely sure about it

Okay so immediately after writing my comment I think I figured it out :
$$aba = b \implies ab = ba^{-1} \text{ and } ba = a^{-1}b$$ so now if I have a string of the form $aabb$ I can write $$ aabb + aba^{-1}b = b a^{-1}a^{-1}b = ba^{-1}ba = bbaa$$ and similar for arbitrary strings so in fact all words are of the form $b^na^m$

 

[anuttarasammyak]

Um yes that is correct but it doesn't really help me, in general you can say that $b^n(x,y) = (x+n,y)$ if n is odd and $(x+n,1-y)$ if n is even but without knowing the structure of the group elements I don't know how to calculate the whole orbit. I think that all members should be of the form $a^nb^m$ but I am not completely sure about it

b^{2n}(x,y)=(x+2n,y)
a and b^2 commute.
a^m b^{2n} (x,y)=(x+2n,y+m)
we observe
ba(x,y)=(x+1,-y)
a^m b^{2n}ba(x,y)=(x+2n+1,-y+m)
We have two sub rectangular lattices.

 

[lavinia]

$a$ and $b^2$ generate a lattice with generators $(0,1)$ and $(2,0)$. This is a free abelian group on two generators.

Conjugating elements of the lattice by $b$ gives another point in the lattice so any element of the group can be written in the form $a^{m}b^{2n}$ or $(a^{m}b^{2n})b$

For instance $ba=(bab^{−1})b$ and $(bab^{−1})$ is an element of the lattice.

So every element of the group is either a pure translation by a lattice point or a lattice translation following the action of $b$.

From this normal form, a direct calculation shows that the action is free.

The group permutes the squares that are bounded by the straight lines parallel to the x and y axes and have integer intercepts. For instance, $b$ shifts the unit square to the adjacent square to the right along the x axis. Note that it does this without fixing any boundary points.

Structurally, the group is an extension of a two dimensional lattice by the group with 2 elements.

$0→L^2→π→Z_2=π/L^2→0$

The element $b$ is a lift of the generator of $Z_2$ to $π$.

Problem: Show that this group is torsion free i.e. it has no elements of finite order.

Problem: Show that the two free abelian subgroups generated by $a$ and $a^{-1}b$ give another description of the group as a split extension of $Z$ by $Z$

$0→Z→π↔Z→0$

where the quotient $Z$ is generated by $a^{-1}b$ and the kernel is generated by $a$. In technical language, $π$ is the semi-direct product of $Z$ with $Z$ in which the action of the quotient $Z$ on the kernel $Z$ by conjugation is multiplication by $-1$.

Question: Does this description also show that the group is torsion free?

Problem: The action of $b$ on the lattice is the identity on $(2,0)$ and negation on $(0,1)$.
Suppose instead that the action was negation of both basis vectors. Would the group still act on the plane freely? Would the group still be torsion free?

Same question if the action of $b$ is the identity on $(2,0)$ and maps $(0,1)$ to $(2,-1)$.

Problem: The quotient space of the plane by the action of the lattice generated by the translations $(0,1)$ and $(2,0)$ is a torus. Show that the action of $b$ on the plane projects to a properly discontinuous involution (map of order 2) of the torus.

Problem: Construct a torsion free extension of the standard 3 dimensional lattice in $R^3$ by the group $Z_2×Z_2$ whose action by conjugation on the lattice is $(x,y,z)→(-x,-y,z)$ and $(x,y,z)→(-x,y,-z)$ and the product $(x,-y,-z)$