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It is known that proving $F_2<SO(3)$ is a key point on the way of demonstrating Banach-Tarski Paradox.
So take two perpendicular rotations $$\left(\begin{matrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{matrix}\right)\text{ and } \left(\begin{matrix}1&0&0\\0&\cos y&-\sin y\\0&\sin y&\cos y\end{matrix}\right),$$ denoting these two rotations by $a,b$ respectively.
In the classical demonstration, $x$ and $y$ is chosen to be $\arccos(1/3)$. $\arccos(1/3)$ is so nice that it is easy to show that $G(x,y):=\langle a,b\rangle\cong F_2$ when $x$ and $y$ are equal to it. The sketch of proof can be found in Wikipedia.
I'm interested in observing what will happen after modifying $x,y$.

Let $G(x,y)=\langle a,b\mid R(x,y)\rangle$, where $R(x,y)$ is the relations. Can we find $R$, or at least find it when $\frac x\pi,\frac y\pi\in\mathbb Q$?

A natural guess is that $R(1,1)=\emptyset$. Also, I noticed that $R(\pi,\pi)=\{a^2,ab,b^2\}$, making $G$ isomorphic to $C_2$.
Also, it is very straightforward to show $\{a^{n_1},b^{n_2}\}\subseteq R(2k_1\pi/n_1,2k_2\pi/n_2)$. I'm not sure about whether it is the full relation when $x,y\in\mathbb Q\pi$.
Knowing that this question requires us to find all relations in a matrix group, I tried to find some paper on this but failed.

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  • $\begingroup$ I think when you ask "can we find..." you are actually asking for an algorithm. It is unclear to be if this is a decidable problem (even under the extra assumptions on rationality that you are making). BTW, I would expect that there are examples of $G(x,y)$ which is not finitely presentable. $\endgroup$ Apr 29, 2020 at 20:05

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