# Subgroups of Mod(S) generated by Dehn twists depend only on intersection numbers?

$$\DeclareMathOperator\Mod{Mod}$$Let $$S$$ be a closed surface and $$\Mod(S)$$ be its mapping class group.

It is a well known fact, proved in the Primer on Mapping class groups for example, that the subgroup of $$\Mod(S)$$ generated by two Dehn twists $$T_a$$ and $$T_b$$ depends only on the geometric intersection number $$i(a,b)$$.

Indeed, if $$i(a,b) \geq 2$$ a ping pong lemma argument on the curve graph shows that $$\langle T_a, T_b \rangle$$ is a free group. Also, if $$i(a,b) = 0$$, then $$\langle T_a, T_b \rangle$$ is clearly free abelian. Lastly, if $$i(a,b) = 1$$, it can be shown that $$\langle T_a, T_b \rangle$$ is a braid group on 3 strands, unless $$S$$ is a torus.

I am wondering whether such a statement can be proven in general, even without knowing a full classification of the groups that can appear. More precisely, if $$\{a_1, \cdots, a_n\}$$ and $$\{b_1, \cdots, b_n\}$$ are two collections of curves such that $$i(a_i,a_j) = i(b_i,b_j)$$ for all $$i,j$$, is it true that $$\langle T_{a_1}, \cdots, T_{a_n} \rangle \simeq \langle T_{b_1}, \cdots, T_{b_n} \rangle$$ as subgroups of $$\Mod(S)$$?.

It seems that the group $$\langle T_{a_1}, \cdots, T_{a_n} \rangle$$ depends only on a regular neighborhood $$N_a$$ of the curves $$a_i$$, which is then homeomorphic to any regular neighborhood $$N_b$$ of the curves $$b_i$$. But in general the injection $$i \colon N_a \to S$$ does not induce an injective map on the mapping class group levels, so this does not seem to give a proper argument.

Any help is greatly appreciated!

Suppose that $$a_1, a_2, a_3$$ all lie in a single handle (surface of genus one, with one boundary component). and all meet exactly once, pairwise. Then the twist about $$a_3$$ lies in the group generated by the others, giving the usual three-strand braid group.
Suppose that $$b_1, b_2, b_3$$ all meet exactly once, pairwise. Let $$N$$ be a regular neighbourhood of their union. Assume that $$N$$ has genus one and three boundary components. Assume that all boundary components are essential in the ambient surface $$S$$. Then the group generated is some Artin group with three generators.