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$$ \newcommand{\Z}{\mathbb{Z}} $$

Consider the group $G=\Z_2\ltimes B^2\Z^2$ where $\Z_2$ acts on $\Z^2$ by interchanging the two factors of $\Z^2$, hence $\Z_2$ also acts on the Eilenberg-MacLane space $B^2\Z^2=K(\Z^2,2)$.

Question: what is the mod 2 cohomology ring $H^*(BG,\Z_2)$ and its module structure over the mod 2 Steenrod algebra?

Here is my attempt (xymatrix failed to compile): There is a fibration sequence $$B^3\Z^2\to BG\to B\Z_2.$$

So we have the Serre spectral sequence $$H^p(B\Z_2,H^q(B^3\Z^2,\Z_2))\Rightarrow H^{p+q}(BG,\Z_2).$$

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    $\begingroup$ This is standard. As a matter of fact you can replace $B^3Z^2$ by $X^2$ (so $X$ will be $K(Z,3)$). Just write $H^*(X^2)=F\oplus T$ where $F$ is free, $T$ is trivial $Z/2$-module, and $E^{\infty}$ looks like $H^0(Z/2,F)\oplus H^*(BZ/2)\otimes T$. I guess normally you can deduce the Steenrod algebra's action just using the Steenrod diagonal, but at worst, the cohomology of the space you are after is detected by $X\times X$ and $X\times BZ/2$. Another way of determining the A-module structure is to consider your space as a stable summand of $QK(Z,3)$ and use the Nishida relations. $\endgroup$
    – user43326
    Commented Nov 8, 2021 at 9:50
  • $\begingroup$ (1) Thanks so much! I voted up. Could you provide a great ref on Nishida relations? $\endgroup$
    – wonderich
    Commented Nov 9, 2021 at 3:30
  • $\begingroup$ (2) What is your $Q$ in the $𝑄𝐾(𝑍,3)$? $\endgroup$
    – wonderich
    Commented Nov 9, 2021 at 3:30
  • $\begingroup$ $Q=\Omega ^{\infty }\Sigma ^{\infty}$ $\endgroup$
    – user43326
    Commented Nov 9, 2021 at 10:19
  • $\begingroup$ Anything that treats the homology of infinite loop spaces will do, for example there is May's book, or this paper people.math.harvard.edu/~dwilson/research/classical-power.pdf by Dylan Wilson $\endgroup$
    – user43326
    Commented Nov 9, 2021 at 10:24

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