# Proving a group with two generators is not free that uses the Brahamagupta-Pell equation

Hello I encountered the following while reading a set of notes on free groups. It's not a homework question.

"Does there exist a rational number $$\alpha$$ with $$0 <|\alpha| < 2$$ such that the group generated by $$\begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 0 \\ \alpha &1 \end{bmatrix}$$ is free.

We mention in passing that several rational $$\alpha$$ with $$0 < |\alpha| < 2$$ are known for which the group generated by $$\begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 0 \\ \alpha &1 \end{bmatrix}$$ is NOT free. The proof interestingly goes via the Brahmagupta-Pell equation."

I was unable to find references that explain this problem and attempted proofs that go through the Brahmagupta-Pell equation. Where may I be able to find some exposition of this? I did contact the author but have not received not yet received a response. If I do, I'll post it.

Thank you for your time and consideration.

* EDIT * Thank you for your comments. Dr. Sapir provided me with information that led me to using additional search criteria. The reference that I found that helps answer the question is in a paper by A.F. Beardon:

"Pell's Equation and two generator free Mobius groups" that is found in the Bull. London Math Soc. 25 (1993) 527-532.

Dr. Beardon considers irrational values of $$\alpha$$. Obviously, I haven't fully read the paper yet but it does provide another reference to a paper by Lyndon and Ullman

"Groups generated by two linear parabolic transformations", Canadian Journal of Math 21 (1969) 1388-1403

that provides insight on when this group can be free for a given rational value of $$\alpha$$.

So, thank you again! I greatly appreciated the assistance! Now I have more enjoyable (and productive) reading to do!

• I don't know what this problem was intending, but, writing $e(\alpha) = \begin{bmatrix} 1&\alpha \\ 0&1 \end{bmatrix}$ and $f(\alpha) = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix}$, we have $\left( e(1/n)^n f(1/n)^{-n} e(1/n)^n\right)^4=\mathrm{Id}$, so there are lots of examples that don't give free groups. Nov 20 '19 at 3:00
• @DavidESpeyer I think your $f(\alpha)$ should have an $\alpha$ in position 2,1 Nov 20 '19 at 3:27
• @WillJagy Is right of course, but I'm past the edit window. Nov 20 '19 at 3:43
• Your quote doesn't look like referring to an "attempted proof thru the BP equation". It says that for some rationals in $]0,2[$ the BP equation can be used to obtain non-freeness.
– YCor
Nov 20 '19 at 8:29
• May I suggest that once you've read the references you have found that you return to summarize them as an answer. Nov 20 '19 at 21:27