On free subgroups in generalized tetraedron groups of type (2, 2, <i>n</i> , 2, 2, <i>m</i> )
The Tits alternative for generalized tetrahedron groups is investigated in the paper. A group <i>G</i> satisfies the Tits alternative, if it contains either a solvable subgroup of finite index or a free group of rank 2. Generalized tetrahedron groups have a presentation of the form Γ = <<i>x</i><sub>1</sub>, <i>x</i><sub>2</sub>, <i>x</i><sub>3</sub>|<i>x</i><sub>1</sub><sup>k1</sup> = <i>x</i><sub>2</sub><sup>k2</sup> = <i>x</i><sub>3</sub><sup>k3</sup> = <i>R</i><sub>12</sub>(<i>x</i><sub>1</sub>, <i>x</i><sub>2</sub>)<sup>m12</sup> = <i>R</i><sub>23</sub>(<i>x</i><sub>2</sub>, <i>x</i><sub>3</sub>)<sup>m23</sup> = <i>R</i><sub>13</sub>(<i>x</i><sub>1</sub>, <i>x</i><sub>3</sub>)<sup>m13</sup> = 1>, where <i>k</i><sub>1</sub>, <i>k</i><sub>2</sub>, <i>k</i><sub>3</sub>, <i>m</i><sub>ij</sub> ≥ 2; <i>R</i><sub>ij</sub>(<i>x</i><sub>i</sub>, <i>x</i><sub>j</sub>) is a cyclically reduced word in the free product <<i>x</i><sub>i</sub>|<i>x</i><sub>i</sub><sup>ki</sup> = 1> ∗ <<i>x</i><sub>j</sub>|<i>x</i><sub>j</sub><sup>kj</sup> = 1>, which is not a proper power. There is a conjecture that these groups satisfies the Tits alternative. At present this conjecture is proved not for all types of generalized tetrahedron groups. In this paper we consider a class of groups of a type Γ = <<i>a</i>, <i>b</i>, <i>c</i>|<i>a</i><sup>2</sup> = <i>b</i><sup>2</sup> = <i>c</i><sup>n</sup> = (<i>ab</i>)<sup>2</sup> = (<i>ac</i><sup>α</sup>)<sup>2</sup> = <i>R</i><sup>m</sup> (<i>b</i>, <i>c</i>) = 1>, где <i>m</i> ≥ 3; <i>n</i> ≥ 6.. The following theorem is proved. Theorem. Let Γ = <<i>a</i>, <i>b</i>, <i>c</i>|<i>a</i><sup>2</sup> = <i>b</i><sup>2</sup> = <i>c</i><sup>n</sup> = (<i>ab</i>)<sup>2</sup> = (<i>ac</i><sup>α</sup>)<sup>2</sup> = <i>R</i><sup>m</sup> (<i>b</i>, <i>c</i>) = 1>, где <i>m</i> ≥ 3; <i>n</i> ≥ 6. Than if Γ is different from the group Γ<sub>1</sub> = <<i>a</i>, <i>b</i>, <i>c</i>|<i>a</i><sup>2</sup> = <i>b</i><sup>2</sup> = <i>c</i><sup>6</sup> = (<i>ab</i>)<sup>2</sup> = (<i>ac</i>)<sup>2</sup> = (<i>bc</i>)<sup>3</sup> = 1>, then Γ contains a non-abelian free subgroup. The group Γ<sub>1</sub> is almost solvable.