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I am trying to understand the description of the group cohomology of $Q_8$ from Adem–Milgram’s “Cohomology of finite groups”. The main result is the following:

Theorem 2.9. In the Lyndon–Hochschild–Serre spectral sequence for $\mathcal Q_8$ where $H = \langle\tau\rangle$, and with $\mathbb F_2$-coefficients, we have $E_2^{i, j} = E_3^{i, j} = \mathbb F_2$ for $i \ge 0$, $j \ge 0$. However, $d_3 \ne 0$. Indeed, $d_3 : E_3^{i, j + 2} \to E_3^{i + 3, j}$ is an isomorphism for $j = 0, 1$. Consequently, $$E_4^{i, j} = E_\infty^{i, j} = \begin{cases} \mathbb F_2 & \text{$i = 0, 1, 2$, $j = 4s, 4s + 1$} \\ 0 & \text{otherwise.} \end{cases}$$ In particular $H^*(\mathcal Q_8; \mathbb F_2)$ is periodic with period $4$, and \begin{alignat*}4 H^0(\mathcal Q_8; \mathbb F_2) & {}= H^3(\mathcal Q_8; \mathbb F_2) && {}= \mathbb F_2, \\ H^1(\mathcal Q_8; \mathbb F_2) & {}= H^2(\mathcal Q_8; \mathbb F_2) && {}= (\mathbb F_2)^2. \end{alignat*} (That the differentials are as stated is clear. Then, note that the $E_4$-term only has classes in the bidegrees stated, and, since the higher differentials go from $(i, j)$ to at least $(i + 4, j - 3)$ there can be no further differentials.)

I understand why all $E_2^{p,q}$ are $\mathbb{F}_2$ , but fail to see why the same holds true for all $E_3^{p,q}$.

Now there is a lemma where the description of the $d_3$ map is given:

Lemma 2.10. In the central extension $$\mathbb Z/2 \xrightarrow\triangleleft \mathcal Q_8 \xrightarrow\pi (\mathbb Z/2)^2$$ the $k$-invariant in $H^2((\mathbb Z/2)^2; \mathbb F_2)$ is $x^2 + y^2 + x y$. In particular $H^*(\mathcal Q_8; \mathbb F_2) = \mathbb F_2[e_4](1, x, y, x^2, y^2, x^2 y = x y^2)$, and $x^3 = y^3 = 0$.

Proof. The three classes $x^2$, $y^2$ and $x^2 + y^2$ all give copies of $(\mathbb Z/4) \times \mathbb Z/2$ as the resulting extensions while the three classes $x y$, $x^2 + x y$, and $x y + y^2$ all give $D_8$. The only remaining non-trivial class is $x^2 + y^2 + x y$, which must, therefore be the $k$-invariant for $\mathcal Q_8$.

Now we consider the spectral sequence associated to the central extension with $E_2$-term $H^*((\mathbb Z/2)^2; \mathbb F_2) \otimes H^*(\mathbb Z/2; \mathbb F_2) = \mathbb F_2[x, y, e]$. We know that $d_2(e) = x^2 + x y + y^2$ determines the $d_2$-differential and $E_3 \cong \mathbb F_2[x, y]//(x^2 + y^2 + x y) \otimes \mathbb F_2[e^2]$. Then the next differential $d_3(e^2) = Sq^1(x^2 + y^2 + x y) = x^2 y + x y^2$, and by comparing with (2.9), there are no further differentials. ▢

I am convinced that $d_2(e)= x^2+y^2+xy$. But I do not understand how one gets the above description of the $E_3$ page and also $d_3(e^2)$. My guess is that it is related to the fact that transgressions commutes with the Bockstein, but it would be of great help if someone explains it in a lucid way.

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    $\begingroup$ For getting the $E_3$-page: you know the $d_2$-differential on generators for $E_2$ as a ring, so you can use the Leibniz rule to propagate the differentials and calculate $d_2$ on everything. Then take cohomology of that cochain complex, and that's $E_3$. Drawing the SS chart in a range of degrees might help to get a feel for what's going on. For the $d_3$ differential on $e^2$, i.e., on $Sq^1(e)$: this is the Kudo transgression theorem, i.e., under appropriate hypotheses, the transgression in the Serre SS commutes with Steenrod operations. $\endgroup$
    – user509184
    Commented Oct 22 at 5:28
  • $\begingroup$ Does “Steenrod operations commute with Transgressions” imply $d_3( Sq^1 (e))= Sq^1 ( d_2(e))$? I am not sure how to apply Kudo’s Theorem since its statement starts with a transgressive element from $E_2^{0,2k}$ (for example, the statement in McCleary), but here $e$ is apparently in $E_2^{0,1}$. Could you please explain this? I am new to all these. $\endgroup$ Commented Oct 22 at 6:10
  • $\begingroup$ Yes, try the version of the Kudo transgression theorem stated in appendix 1, Theorem A1.5.7, of Ravenel's book "Complex cobordism and stable homotopy groups of spheres." It is for the spectral sequence of an extension of Hopf algebroids, rather than the Serre SS of a fiber sequence, but the Lyndon-Hochschild-Serre SS you are running is a special case of both. $\endgroup$
    – user509184
    Commented Oct 22 at 6:27
  • $\begingroup$ Thanks a lot, this helps. However could you elaborate how $E_2 \cong \mathbb{F_2}[x,y,e]$ implies $E_3 \cong \mathbb{F_2}[x,y]/ (d_2(e)) \otimes \mathbb{F_2}[e^2]$ $\endgroup$ Commented Oct 22 at 8:09
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    $\begingroup$ Whenever possible, it is better to use (searchable) text than (un-searchable) images. I have transcribed your images as best as I could. Once you have checked that the transcription is correct, and/or fixed any errors, you can delete the images. $\endgroup$
    – LSpice
    Commented Oct 22 at 17:44

1 Answer 1

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I am converting my comments on this question into an answer because I see that they do answer the question.

  • For getting the $E_3$-page: you know the $d_2$-differential on the generators $x,y,e$ for $E_2$ as a ring, so you can use the Leibniz rule to propagate the differentials and calculate $d_2$ on everything in $E_2$. In particular, this gives you $d_2(x^i y^j e^k)$ for all $i,j,k$. I think you will very quickly find that the cocycles are the monomials with $k$ even, and the coboundaries are precisely everything divisible by $d_2(e)$. Hence the $E_3$-page consists of a polynomial algebra on $x$, $y$, and $e^2$, modulo the ideal generated by $d_2(e)$.

  • For the $d_3$-differential on $e^2$, i.e., on $Sq^1(e)$: this is the Kudo transgression theorem, i.e., under appropriate hypotheses, the transgression in the Serre SS commutes with Steenrod operations. There is a nice writeup of the Kudo transgression theorem in appendix 1, Theorem A1.5.7, of Ravenel's book "Complex cobordism and stable homotopy groups of spheres." It is for the spectral sequence of an extension of Hopf algebroids, rather than the Serre SS of a fiber sequence, but the Lyndon-Hochschild-Serre SS of a group extension is a special case of both.

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