TITLE: Subgroups of the mapping class group of a surface generated by Dehn twists
QUESTION [8 upvotes]: Let $G$ be the mapping class group of a surface of genus $g > 1$. Is it known for which positive integer $k$ one can find a subgroup $H$ of $G$ generated by a finite number of Dehn twists and a Dehn twist $\tau \in G$ which is not in $H$ but such that $\tau^k$ is in $H$, $k$ being minimal for this property ?
Using the chain relation, one can construct examples with $k = 2$, but I cannot find any examples with higher powers.
EDIT: I realized I should put a little motivation to this question. So let us take $U_d$ the space of smooth degree $d$ planar complex curves. There is a monodromy map $\pi_1(U_d)\rightarrow G$, and the image of this map is generated by Dehn twists obtained for example by taking a generic pencil of curves. Now I can construct a loop in $U_d$ whose monodromy is a power of some Dehn twist and I am wondering if there should be a reason coming from the structure of $G$ for the Dehn twist itself to be in the image of the monodromy.

REPLY [3 votes]: Edited: 
The paper Dehn twists have roots proves that Dehn twists have roots, naturally.  The limits on that construction can be found in the paper Roots of Dehn twists; they bound the degree of the root (linearly, as I recall) in terms of the complexity of the surface.  
Now, if I understand correctly, you are asking for ways to realize $\tau_\alpha^p$, a power of a Dehn twist about $\alpha$, as a product of other twists.  If we found a root of $\tau_\alpha^p$, which was not itself a twist, then that would answer your question.  But I think that the main idea behind the above papers will work.  For: any root of $\tau_\alpha^p$ has the same canonical reduction system, namely $\alpha$.  So the root of $\tau_\alpha^p$ stabilizes $\alpha$.  You can now build a periodic map in the complement of $\alpha$ that does a fractional twist about $\alpha$: for example $p/q$ fraction of a right twist.  Take the $q$-power of this get the desired $\tau_\alpha^p$.
Note that it is far from clear that these constructions are the only way to get a subgroup $H$ of the type you desire.