A Uniform Bound for Free Pairs of Bicyclic Units and for Free Points at Algebraic Integers

Gonçalves and del Río asked two questions in the problem session of Oberwolfach Report 55/2007 (Problem 28); the first is also the first half of Question 5.4 of their paper in J. Group Theory 11 (2008). A nonzero complex number \(z\) is a free point if the matrices \(\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\) and \(\left(\begin{smallmatrix}1&0\\z&1\end{smallmatrix}\right)\) generate a free group, and \(m(z)\) is the least positive integer \(k\) for which \(kz\) is a free point. For a finite non-Hamiltonian group \(G\), \(m(G)\) is the least positive integer \(m\) such that \(b_1\) and \(b_2^m\) generate a free group whenever \(b_1,b_2\) are bicyclic units of \(\mathbb{Z} G\) that generate a group which is not nilpotent. Is \(m(z)\) bounded over the nonzero algebraic integers \(z\), and is \(m(G)\) bounded over all such groups \(G\)? We observe that both are bounded by \(4\). A nonzero algebraic integer has a Galois conjugate of modulus at least \(1\), freeness of a point is invariant under Galois conjugation, and every complex number of modulus at least \(4\) is a free point (Sanov, Brenner); hence \(kz\) is a free point for every integer \(k\) with \(|k|\ge4\), and \(m(z)\le4\), with equality for \(z=\pm1\). For bicyclic units \(1+a\) and \(1+b\), the element \(ab\) acts on \(\mathbb{Z} G\) by an integer matrix, so if it is not nilpotent it has an eigenvalue \(\lambda\) with \(|\lambda|\ge1\), and the two-dimensional reduction of Gonçalves and del Río turns \((1+a,(1+b)^m)\) into \(\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\), \(\left(\begin{smallmatrix}1&0\\m\lambda&1\end{smallmatrix}\right)\); hence \(m(G)\le4\), also for the variant \(M(G)\) that allows units of different types. The result is an elementary combination of known facts. Exact computations confirm \(m(S_3)=1\), \(M(S_3)=2\) and \(m(S_4)=2\) and give \(M(S_4)=2\); exhaustive computations with numerically located roots give \(m(S_5)=M(S_5)=2\). Whether \(m(G)\le2\) for all \(G\), the second part of Question 5.4, remains open. No answer was found in the literature accessible to us; a search that finds nothing is not a proof of novelty, and four later sources could not be read. This is an unrefereed note.