Free subgroups of $\mathrm{GL}(2,\mathbb{Z})$

Question

Is there a bound $B$ such that every 2-generator subgroup $G = \langle a, b \rangle \le {\rm GL}(2,\mathbb{Z})$ whose generators do not satisfy a relation of length $\leq B$ is free?

If it exists, such bound must be at least 18, as the example $$ G = \left< \left( \begin{array}{rr} 5 & 4 \\\ -1 & -1 \end{array} \right), \left( \begin{array}{rr} 6 & 1 \\\ -1 & 0 \end{array} \right) \right> $$ shows: the shortest relation satisfied by the generators $a$ and $b$ is $a^{-2}b(ab^{-1})^3a^2b^{-1}(a^{-1}b)^3 = 1$.

Remarks:

• Obviously the question can be generalized to $m$-generator subgroups of ${\rm GL}(n,\mathbb{Z})$.

• The crystallographic restriction gives a positive answer to case $m = 1$ of the above generalization, thus our case $m = n = 2$ is the minimal case which is not covered.

• By the Tits alternative, a subgroup of ${\rm GL}(n,\mathbb{Z})$ has either a free subgroup or a solvable subgroup of finite index.

Answer 1

In Olʹshanskiĭ, A. Yu.; Sapir, M. V. On $F_k$-like groups. (Russian) Algebra Logika 48 (2009), no. 2, 245--257, 284, 286--287; translation in Algebra Logic 48 (2009), no. 2, 140–146, we proved (Theorem 2) that every non-virtually cyclic hyperbolic group is $F_k$-like, that is for every $m$ it has a generating set consisting of $k$ elements which do not satisfy any relation of length $\le m$. The minimal $k$ that works is 1 plus the number of generators of the group. In particular, for $GL_2(Z)$, $k=3$.

For the group $\mathbb{Z}*\mathbb{Z}_2$ (as in Yves' comment), $k=2$ also works: one can adapt the proof of Theorem 1 of the paper (we prove, in particular, that every group with 2-generated presentation satisfying $C'(1/6)$ is $F_2$-like).