Calculating cohomology group $H^3(point group,\mathbb{Z})$ using GAP program I'm trying to compute $H^3(point group,\mathbb{Z})$ for all the 32 point groups in 3D which has some applications in physics. Unfortunately, I could not find literature discussing this problem. So I tried to use GAP program to compute it. 
I used the following code to do the computation:
gap> GroupCohomology(PointGroup(SpaceGroup(3,x)),3);

which is basically calculating the point group corresponding to the space group No.x. I have to admit it might be a bit inefficient because I'm new to GAP program. 
My problem is that when I take $x=207$, because I want to obtain $H^3(O,\mathbb{Z})$ where $O$ is the octahedral group, some error props out. And when I try $x=210$, which corresponds to the same group $O$, it took too long and did not give me any result.
My ultimate goal is to obtain the cohomology group of point group $O$. Can anyone give me the result(by some analytical method or citing from some literature) or simply help me with the GAP program to get the result?
Edit: The octahedral group $O$ (symmetry group of an octahedron with only orientation preserving elements) is isomorphic to $S_4$, therefore it is in fact very easy to deal with.
 A: The Kunneth formula for group cohomology is:
$$H^n(G_1 \times G_2; \Bbb Z) \cong 
\bigoplus_{i= 0}^n H^i(G_1;\Bbb Z) \otimes_{\Bbb Z} H^{n-i}(G_2;\Bbb Z) 
\oplus\bigoplus_{p =0}^{n+1} \text{Tor}^{\Bbb Z}(H^p(G_1;\Bbb Z),H^{n+1-p}(G_2;\Bbb Z))
$$
Most of these terms will be $0$ due to the fact that there are a lot of cyclic group cohomologies floating around. For the $n=3$ case we get an expression which relies only on $H^n$ for $n=0, \dots, 4$ of each direct product factor. So to compute $H^1(G_1 \times G_2)$ (assuming $\Bbb Z$ coefficients here) we need to compute 10 cohomology groups, 5 for each factor.
You can start to see why GAP was invented. Anyway since the entries for Tor are always abelian groups they should be pretty easy to calculate. (They will distribute over direct sums of groups etc.)
Now the idea is to keep doing this until you have cohomology groups you know. All groups in this table have factors whose cohomology groups are known.
This calculation is tedious so once your done write it down somewhere safe.
Alternatively you could get GAP to do it but I do not know how. This is sort of the brute force way of getting what you want.
One more thing, I would try and factor groups with multiple factors so that a single cyclic group is your second factor as this will naturally cancel many of your terms. Also if you do not know about tensor products of abelian groups Google is your friend.
Standard references for this kind of work would probably be Group cohomology by Brown but that is still not what you are looking for probably.
