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

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.

• The answer should be no. If $k$ is large enough ($k>B$ probably is OK) then in the free group $\langle a,b\rangle$, no nontrivial word of length $\le B$ is a consequence of $(ab^k)^2$ (well, there is something to check). Since $ab^k$ is primitive, the quotient is the free product $C*C_2$ where $C$ is cyclic infinite and $C_2$ is cyclic of order 2. Thus $C*C_2$ has a generating set for which the $B$-ball has no nontrivial relation. Since it embeds into $GL_2(\mathbf{Z})$, it should answer the question.
– YCor
Commented Jan 25, 2013 at 21:35

## 1 Answer

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).