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I only know through stories that mod 3 moore spectrum is not associative. I do not know of any proof. I have been informed that Toda had proved it in the paper "Extended $p^{th}$ power". I was not able to follow it. Can anybody give me a proof that mod 3 Moore spectrum is not associative?

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This is also Lemma 6.2 in Toda's On spectra realizing exterior parts of the Steenrod algebra (, which you might find helpful. (I don't know if this proof is any different from the one in the Toda reference you mention, which I don't have available.) –  Eric Peterson Jan 15 '13 at 20:47
That paper of Toda's is a little concentrated. The nature of the answer to this question will probably depends pretty heavily on whether you know what a Massey product is. You can show that there must be a map from $M$ to $H\mathbb{Z}/3$ which preserves the unit and multiplication. The resulting map $H_* M \to H_* H\mathbb{Z}/3$ has as image a square-zero class in the dual Steenrod algebra ($\tau_0$) whose triple Massey product $\langle \tau_0, \tau_0, \tau_0 \rangle$ is not in the image. –  Tyler Lawson Jan 16 '13 at 4:42
Hi @TylerLawson. Could you say a bit more about the shape of the argument? You are saying that if $M$ had an $A_{\infty}$-structure, and the above map is induced from an $A_{\infty}$-map then for all elements in the image, their triple Massey products (a representative of which is given by the $A_{\infty}$-structure) must always be in the image because - is this correct? –  Elden Elmanto Jul 11 '14 at 20:39

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