# A subgroup generated by two noncommuting elements in a group

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? –  Calvin Lin Dec 31 '12 at 11:53
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. –  Teddy Dec 31 '12 at 11:59
Why not take $RS^1\times \mathbb C$ with $(a_1,b_1)\cdot (a_2,b_2) = (a_1 a_2, a_1b_2+b_1)$? –  Hagen von Eitzen Dec 31 '12 at 11:59
@HagenvonEitzen You're right. I actually over complicated the problem for no reason. I'll withdraw my question, and think it over. –  Teddy Dec 31 '12 at 12:09

## 1 Answer

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: $$(0,1)^{(-2,1)}=(0,-1)=(0,1)^{-1}$$ - My apologies. I changed the notation after the answer was posted. In the new notation, DonAntonio's answer is: $$(a,b)^n=\begin{cases}\left(a^n\;,\;\frac{a^n-1}{a-1}b\right)&,\;\; a\neq 1\\{}\\{}\\{.}\\{.}\\\(0,nb)&,\;\;\,a=1\end{cases}$$ – Teddy Dec 31 '12 at 13:00 Why did you do that? It probably would have been better to leave the original question (which was pretty interesting imfho) and open a new thread, wouldn't it? And the upper lime in your comment above is wrong: that is not what is written in my answer. – DonAntonio Dec 31 '12 at 13:03 Thanks for your reply. You mean the lower line should be$(1,nb)$, correct ? Again, I'm sorry for pulling the notation under your answer. – Teddy Dec 31 '12 at 13:12 No, it is$\,(0,1)^n=(0,nb)\,\$ , according to the first definition of the product you gave. –  DonAntonio Dec 31 '12 at 13:15