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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).

Hans
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    This looks like a good example. – Pedro Jun 28 '17 at 07:49
  • I have read that answer (and the corresponding exercises in Brown's book), but I am still unsure about the computation of the map $\tau$. In the answer you linked, he refers it as "the" chain map without further comments, which implies (?) some kind of uniqueness (up to homotopy?). On top of that I do not really get how he computes it. For even more confusion, Weibel uses another chain map to compute the exact same action and just says "look at this map and compute" which seems a bit artificial to me (I mean, it is rigorous, but I don't see how to come with such a map). – Hans Jun 28 '17 at 07:57
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    You find that map by looking for it: start by building it by hand and hope you see the pattern. That's part of the art of computing this sort of things. – Mariano Suárez-Álvarez Jun 28 '17 at 08:14
  • You should read a book describing how to get all the differentials. In fact, evert textbook which constructs the spectral sequence associated to a double complex does. – Mariano Suárez-Álvarez Jun 28 '17 at 08:15

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