Mapping Class Group

Question

$\newcommand{\MCG}{\mbox{MCG}}$Let $\alpha$, $\beta$ be non-isotopic, non-separating curves on a surface $S$ (meaning that "cutting along " them will not disconnect the surface). How do we show that the Dehn twists $D_{\alpha} , D_{\beta}$ are conjugate in $\MCG(S)$ , but not isotopic to each other? I know that the mapping class group is not Abelian, and that $\MCG(S)$ is generated by Dehn twists about essential curves ($2g+1$ twists as the best possible when S has empty boundary). Maybe we can show (after showing that the twists are conjugate in $\MCG(S)$) that conjugate curves in a non-Abelian curves cannot be isotopic? I have been going over Casson and Bleiler's book, but I can't find an answer. So, I guess to show $D_{\alpha}, D_{\beta}$ are conjugate , we need to find a curve $\gamma$ so that $$D_{\alpha}=D_{\gamma}^{-1} D_{\beta} D_{\gamma} .$$

I don't know enough about $\MCG(S)$ to figure out how to do this. Also, I cannot see how to show $D_{\alpha}, D_{\beta}$ are non-isotopic. Maybe because both are (class-wise) generators and different generators are not isotopic?

Any ideas, please?

Answer 1

In general there is no such curve $\gamma$. One way to find the conjugating map is to use the classification of surfaces. (That is, use the fact that $S - \alpha$ is homeomorphic to $S - \beta$.) Another approach is given in Lickorish's famous paper "A representation of orientable combinatorial 3-manifolds". He uses the fact that if $\alpha$ and $\beta$ intersect once then $D_\alpha D_\beta$ throws $\alpha$ onto $\beta$. He also proves that the "graph of non-separating curves" - where vertices are non-separating curves and edges are intersection number one - is connected. (I believe that this proof was also known to Dehn.)

[EDIT]

As for your second question: This is Fact 3.6 of the "Primer on mapping class groups" by Farb and Margalit. There are other proofs of this, but I'll guess that any proof eventually comes down to the Bigon Criterion.