I am looking for $H^*(BZ/2p , Z/2p)$ where $p$ is odd prime.We can calculate cohomology groups by using gysin exact sequence and universal coefficient theorem.But I am unable to calculate the ring structure.looking for help. Thanking you.

  • $\begingroup$ Have you tried using a spectral sequence? $\endgroup$ – JHF Mar 26 '15 at 23:20
  • $\begingroup$ @JHF: You mean using the serre fibration $K(Z,1) \rightarrow K(Z,1) \rightarrow K(Z/2p ,1)$,where first map is multiplication by 2p. $\endgroup$ – Math Mar 26 '15 at 23:26
  • $\begingroup$ @JHF :It is easy to see that $H^{2i}(BZ/2p , Z/2p) =Z/2p\lbrace t^i \rbrace$ where $t$ belongs to $H^2(BZ/2p,Z/2p)$.By SSS we guarantee that there is class $s$ belongs to $H^1(BZ/2p , Z/2p)$ and $s^2$= 0 because otherwise it will survive at $\infty$-page and that contradict to cohomology of $K(Z,1)$.Is it right? $\endgroup$ – Math Mar 26 '15 at 23:44
  • $\begingroup$ You may be right, but how did you differentiate between the cases $s^2 = 0$ and $s^2 = t$ (or some other multiple of $t$)? $\endgroup$ – JHF Mar 27 '15 at 1:00
  • $\begingroup$ @JHF:You are right. Can u settle it for me?Can you tell me the explicit ring structure? $\endgroup$ – Math Mar 27 '15 at 2:40

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