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Let the "rational unit circle" be $ RS^1 = \{ e^{i \theta \pi} \, |\, \theta \in \mathbb{Q}\}.$

Let $G$ be the group of linear functions, $ G = \{\varphi(z)=az+b \, | \, a\in RS^1, b\in \mathbb{C} \},$ (that is, linear functions which are "rational"-rotation + translation) where multiplication is function composition.

Clearly, we may view $G$ as the set of pairs $ G = \{ (a,b) \, | \, a\in RS^1, b\in \mathbb{C}\},$ and define multiplication as $ (a_1,b_1) \cdot (a_2,b_2) = (a_1 a_2, a_1 b_2 + b_1).$

Question: Suppose $\varphi_1,\varphi_2\in G$ do not commute. What is the subgroup they generate.

Any comments regarding the structure of this group are welcome.

P.s. In a previous formulation of the problem, I defined $ G = \{ (a,b) \, | \, (a+1)\in RS^1, b\in \mathbb{C}\},$ with multiplication: $ (a_1,b_1) \cdot (a_2,b_2) = (a_1 a_2 + a_1 +a_2, a_1 b_2 + b_1 +b_2).$ DonAntonio's answer below refers to this older formulation.

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    @HagenvonEitzen You're right. I actually over complicated the problem for no reason. I'll withdraw my question, and think it over.2012-12-31

1 Answers 1

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It is easy, and only slightly messy, to show by induction that

$(a,b)^n=\begin{cases}\left((a+1)^n-1\;,\;\frac{(a+1)^n-1}{a}b\right)&,\;\; a\neq 0\\{}\\{}\\{.}\\{.}\\(0,nb)&,\;\;\,a=0\end{cases}$

Perhaps this helps a little and later, maybe, I'll add something else

Added: Let us take for example the non-commuting elements $\,(-2,1)\,,\,(0,1)\,$:

$(-2,1)^2=\left((-1)^2-1\;,\;\frac{(-1)^2-1}{-2}\right)=(0,0)\Longrightarrow \mathcal Ord(-2,1)=2$

and since clearly

$\mathcal Ord(0,1)=\infty$

We have that $

$\langle\,(-2,1)\,,\,(0,1)\,\rangle\cong\Bbb Z\rtimes C_2=\,\text{infinite dihedral group}$

with the inversion action:

$(0,1)^{(-2,1)}=(0,-1)=(0,1)^{-1}$

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    No, it is $\,(0,1)^n=(0,nb)\,$ , according to the first definition of the product you gave.2012-12-31