Subgroup of mapping class group generated by two Dehn twists Let $S$ be a surface (possibly with boundary, and punctures), and let $\alpha,\beta$ be two simple closed curves on $S$ which intersect once.  If $a,b$ denote the isotopy classes of $\alpha,\beta$, respectively, then why is the subgroup of $\text{Mod}(S)$ generated by $T_a,T_b$ isomorphic to the braid group $B_3$? [Here, $T_a$ is the Dehn twist around $\alpha$, and $T_b$ the Dehn twist around $\beta$.] I understand why the relation $T_aT_bT_a=T_bT_aT_b$ holds, but why is this the only relation? 
If it makes any difference, I am reading the "Primer on Mapping Class Groups" by Farb and Margalit (available here); they claim this is true, but give no proof. The relevant section in that PDF is 3.5, specifically pages 91--94 (in the PDF).
Thanks for any help,
Steve
 A: A regular neighborhood of these two curves is a 1-holed torus, so you're question is equivalent to asking why the mapping class group $M_{1,1}$ of a 1-holed torus is isomorphic to the 3-strand braid group $B_3$.  The key observation is that you can construct a homomorphism $f : M_{1,1} \rightarrow B_3$ as follows.  Let $i : \Sigma_{1,1} \rightarrow \Sigma_{1,1}$ be the hyperelliptic involution.  This is not an element of $M_{1,1}$ since it rotates the boundary component by $\pi$, but it lies in a degree 2 extension $M_{1,1}'$ of $M_{1,1}$.  In fact, $i \in M_{1,1}'$ lies in the center of $M_{1,1}'$.  It follows that $M_{1,1} \cong M_{1,1}' / \langle i \rangle$ acts (modulo homotopy) on the quotient $\Sigma_{1,1}/i$, which is a disc with $3$ punctures.  Since $B_3$ is the mapping class group of a $3$-punctured disc, this give you a homomorphism $f : M_{1,1} \rightarrow B_3$.  The dehn twists $T_{\alpha}$ and $T_{\beta}$ go to the standard generators of $B_3$, so the proof is completed by observing that they satisfy the braid relation.
