Let μ be a real number. The Möbius group Gμ
is the matrix group generated by
formula here
It is known that Gμ is free if [mid ]μ[mid ] [ges ] 2
(see [1]) or if μ is transcendental (see [3, 8]).
Moreover, there is a set of irrational algebraic numbers μ which is dense in
(−2,2) and for which Gμ is non-free [2, p. 528].
We may assume that μ > 0, and in this paper we consider rational
μ in (0, 2). The following problem is difficult.
formula here
Let [Gscr ] denote the set of all rational numbers μ in (0, 2) for which
Gμ is non-free. In 1969 Lyndon and Ullman
[8] proved that [Gscr ]nf contains the elements of the forms
p/(p2 + 1) and 1/(p + 1), where
p = 1, 2, …, and that if μ0 ∈ [Gscr ]nf
then μ0/p ∈ [Gscr ]nf for
p = 1, 2, …. In 1993 Beardon [2]
studied problem (P) by means of the words of the form
ArBsAt
and ArBsAtBuAv, and he obtained a sufficient condition for
solvability of (P), included implicitly in [2, pp. 530–531],
by means of the following Diophantine equations:
formula here