2-generator subgroups of an Artin group of small type

Suppose I have an Artin group $G$ of small-type, meaning that the generators either commute or braid. E.g a braid group. Take two generators $g, h$ and arbitrary conjugates of these generators $xgx^{-1}$ and $yhy^{-1}$. Is it true that the only possibilities for the subgroup $$\langle xgx^{-1},yhy^{-1}\rangle \cong\langle g, (x^{-1}y)h(x^{-1}y)^{-1}\rangle$$ are $\mathbb Z^2$, the braid group $B_3$, and the rank 2 free group $F_2$ (or $\mathbb{Z}$ in the degenerate case when $g=h$)?

• Is the similar statement for all Coxeter groups (the subgroup $\langle g, h^x\rangle$ is either free or isomorphic to a visible subgroup of $G$) true? How about RACGs? – user6976 Feb 14 '18 at 18:33
• In the Coxeter setting a similar statement indeed holds and is easy to prove. – Misha Feb 14 '18 at 19:18
• This is also true for RAAGs. If $x, y, z$ are generators of a RAAG $G$, then the subgroup generated by $x$ and $z^{-1}yz$ is also a RAAG. In fact, it is a parabolic subgroup of the RAAG which is the kernel of the map $G \to \mathbb{Z}/2$ which maps $z$ to $1$ and every other generator to $0$. (This kernel is isomorphic to the RAAG defined by doubling the original defining graph along the star of $z$; new vertices correspond to conjugates of the original vertices.) It then follows by induction on the length of an element $g$ that the subgroup generated by $x$ and $g^{-1}yg$ is a RAAG. – Robert Bell Feb 15 '18 at 19:18
• Am I correct in thinking that it Is unknown whether the word problem is solvable in Artin groups of small type? – Derek Holt Feb 16 '18 at 8:26
• @DerekHolt You are correct. It's only known in very few select cases. We don't even know if the Artin group of small type defined by the complete bipartite graph $K_{5,1}$, a 5-pointed star, has solvable word problem. – Harry Reed Feb 16 '18 at 8:30