Are mapping class groups of orientable surfaces good in the sense of Serre? A group G is called ‘good’ if the canonical map $G\to\hat{G}$ to the profinite completion induces isomorphisms $H^i(\hat{G},M)\to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference.
I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $\mathbb{Z}/n\mathbb{Z}$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess.
So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good?
Thanks for any help.
 A: The braid groups are good (which are mapping class groups of punctured disks) by Proposition 3.5 of 
Grunewald, F.; Jaikin-Zapirain, A.; Zalesskii, P. A., Cohomological goodness and the profinite completion of Bianchi groups., Duke Math. J. 144, No. 1, 53-72 (2008). ZBL1194.20029. (You can find a version here but missing the statement of this Proposition.)
The summary of the proof is that if $G$ is good, and $H$ is commensurable with $G$, then $H$ is also good (Lemma 3.2). Hence we may pass to the pure braid group $P_n$. This is an extension of $P_{n-1}$ by $F_{n-1}$ (essentially the Birman exact sequence). Hence by induction, $P_n$ is good using Lemma 3.3 (residually finite extensions of good groups by certain good groups such as finitely generated free groups are also good). 
By a similar inductive argument, the mapping class groups of the $n$-punctured sphere and torus are good. Moreover, the mapping class group of the genus $2$ surface is a central extension of the mapping class group of the $6$-punctured sphere by $\mathbb{Z}/2$ (generated by the hyperelliptic involution). Hence by Lemma 3.3 the mapping class group of the genus $2$ surface (and also with $n$ punctures by induction) is good.   
Mapping class groups of surfaces with boundary are central extensions of the (pure) punctured surface mapping class groups, so by similar reasoning they are good if the punctured ones are. Thus one sees that the general case would follow from the closed case.
A: This is an open and probably very difficult question.  There have been purported proofs (for instance, this one), but they have all had fatal flaws.
The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian).  For instance, it contains huge numbers of nonabelian free subgroups.  One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free.  See
Ishida, Atsushi,
The structure of subgroup of mapping class groups generated by two Dehn twists.
Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241. 
