Presentation of $H^2(\overline{M}_{0,n},\mathbb{Z})$ as an $S_n$-module? Let $\overline{M}_{0,n}$ be the moduli space of genus zero curves with $n$ marked points. Let $I=\{\{S,S^c\}|S\subset\{1,\dots,n\},|S|\geq2,
|S^c|\geq2\}$ be the set of partitions of $\{1,\dots n\}$ into two subsets, each has at least two elements.
Keel shown that the cohomology group $H^2(\overline{M}_{0,n},\mathbb{Z})$ is generated by boundary divisor classes $\delta_{\{S,S^c\}}$, where $\{S,S^c\}\in I$.
(Here $\delta_S$ is the divisor in $\overline{M}_{0,n}$, consisting of curves with two components, marked by $S$ and $S^c$, and their further degenerations.)
Thus we have a presentation $$\mathrm0\to K\to\bigoplus_{\{S,S^c\}\in I}\mathbb{Z}\cdot\delta_{\{S,S^c\}}\to H^2(\overline{M}_{0,n},\mathbb{Z})\to 0.$$ 
Keel shown that the kernel $K$ is generated by equations $$\sum_{i,j\in S;k,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;j,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,l\in S;k,j\notin S}\delta_{\{S,S^c\}},$$ for any four distinct elements $\{i,j,k,l\}\subset\{1,\dots,n\}$. These give $2\cdot{n\choose 4}$ such equations. But the rank of $K$ is only $\frac{n(n-3)}{2}$. ($\#I=2^{n-1}-n-1$, $\mathrm{rank}H^2(\overline{M}_{o,n},\mathbb{Z})=2^{n-1}-{n\choose2}-1$), so these relations are very much dependent..
The question is, would there be a good presentation of $K$, my goal is to calculate the group cohomology $$H^1(S_n,H^2(\overline{M}_{0,n},\mathbb{Z}))?$$
 A: This is a bit too long for a comment.
If you take the relation
$$
\sum_{i,j\in S;k,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;j,l\notin S}\delta_{\{S,S^c\}}
$$
and add to both sides $\sum\limits_{i,j,k\in S;l\notin S}\delta_{\{S,S^c\}}$, you get
$$
\sum_{i,j\in S;l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;l\notin S}\delta_{\{S,S^c\}}.
$$
Of course, this is totally reversible, so you can take these elements instead. (This essentially recovers the presentation of cohomology of $\overline{M}_{0,n}$ when this space is constructed as a De Concini-Procesi wonderful model of the Coxeter hyperplane arrangement of type A.)
Now we note that for fixed $i,l$, it is enough to choose $n-3$ equations among the $(n-2)(n-3)/2$ equations indexed by different choices of $j,k$ to force all of them to hold, and the symmetric group action on those seems rather straightforward; I suspect that this would be quite useful for your purposes.
[Things would be even simpler if you could fix $l$ once and forever (that is, restrict to $S_{n-1}$ inside $S_n$), but I suspect you do not want to do that.]
