I will denote by $Q_8$ the quaternion group. According to this page: Groupprops wiki, the cohomology seems a bit tricky (at least, that is what I deduce from the multiple question marks '?') so I will only describe my problem in the case of homology.
Using the Lyndon-Hochshild-Serre spectral sequence, I have two different approaches in mind given with two different choices of a normal subgroup: either $\{1,-1,i,-i\}$ or $\{1,-1\}$. The former gives a better looking quotient $\Bbb Z/2$ while the former is the center of $Q_8$ and results in a trivial action on the homology group.
Let's assume that I picked $K=\{1,-1,i,-i\} \lhd Q_8$. From what I understood (I have never really computed actions on homology groups), if I want to compute the action $(c_g)_*$ induced on homology by the conjugation with an element of $g \in Q_8/K$, I have to apply the following recipe:
- pick $F_\bullet$ a projective resolution of $\Bbb Z$ over $\Bbb ZK$
- find a chain map $\tau : F_\bullet \rightarrow F_\bullet$ such that $\tau(kx) = c_g(k)\tau(x)$ for all $k \in K$ (where $c_g(k) = gkg^{-1}$)
- Realize that $F_\bullet \otimes_K \Bbb Z \rightarrow F_\bullet \otimes_K \Bbb Z$ is given by $x\otimes y \mapsto \tau(x) \otimes gy = (c_g)_* (x\otimes y)$
- Finally, finding such a $\tau$ should give me the expression of $(c_g)_*$, which is the action I was looking for.
Is this procedure correct? If yes, how does one find such a chain map in general (I mean, in the most common cases)?
Finally, after trying the two approaches, it seems that my $E_2$ page is filled with a lot of non-zero terms, how does one compute the actual differentials to get to the $E_3$ page? I have not read any book describing how to get the differentials (most of them stop at the $E_2$ page).