@PiotrPstrągowski That's strange indeed.

@SaalHardali The trick is that there aren't that many K(n)-modules

I think what's strange here is that there's no distinction between right and left...

I'll need to read the precise statement to really understand what it means for p=2

I think the point is that the π_*K(n)-bimodule structure on π_*K(n) is symmetric, and every K(n)-module is free

@DenisNardin I'm so confused. I'm not sure what exactly needs to be checked in order for the functor to $K(n)_{\ast}$-module to be symmetric monoidal. What you said right now sounds like it would imply that the Kunneth isomorphism is automatically symmetric. Before I say anything further I have to look carefully at the bimodule action of $K(n)_{\ast}$ on $K(n)_{\ast} K(n)$