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Mar 1 at 10:47 comment added David Roberts In a comment in the original posting, @Yuli (which was converted to a comment, as it wasn't an answer to the question), Dave Benson said "Welcome to Math Overflow!"
Feb 29 at 21:52 comment converted from answer Yuli Rudyak I am Rudyak. I only recently saw this discussion and regret not seeing it sooner. Anyway, it's nice. Special thanks to Nick Kuhn.
Mar 16, 2018 at 21:12 history edited Nicholas Kuhn CC BY-SA 3.0
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Mar 16, 2018 at 8:12 vote accept Chi Cheuk Tsang
Mar 16, 2018 at 8:12 comment added Denis Nardin @Tsang You can find computations of this kind in Adams' blue book (Generalized cohomology and stable homotopy).
Mar 16, 2018 at 7:09 comment added Chi Cheuk Tsang Great answer! Just one thing: can you please explain how we get $H^*(H \mathbb{Z};\mathbb{Z}/p)$ and $H^*(H \mathbb{Z}/p^m;\mathbb{Z}/p)$ from $H^*(H \mathbb{Z}/p;\mathbb{Z}/p)=A$? I tried using a cofibre sequence like $H\mathbb{Z} \to H\mathbb{Z} \to H\mathbb{Z}/p$ but I couldn't get it to work.
Mar 16, 2018 at 3:41 history answered Nicholas Kuhn CC BY-SA 3.0