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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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    Out of curiosity, why is this done for the rational unit circle, instead of the entire circle?2012-12-31
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    It may be extended to the whole unit circle; it will still be a group. I'm only interested in the rational case for a problem I'm trying to solve.2012-12-31
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    Why not take $RS^1\times \mathbb C$ with $(a_1,b_1)\cdot (a_2,b_2) = (a_1 a_2, a_1b_2+b_1)$?2012-12-31
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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

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