# $QS^0$ isn't a product of Eilenberg-Mac Lane

I am reading Lewis' paper "Is there a convenient category of spectra?". To prove the main result on the non-existence of such a nice category, he shows that otherwise the unit component of $QS^0= \varinjlim \Omega^n S^n$ would have to be weakly equivalent a product of Eilenberg-Mac Lane spaces. So far so good, but it isn't immediately clear to me:

Q: Why can't the unit component of $QS^0$ be a product of Eilenberg-Mac Lane spaces?

This should probably be very easy, given how the paper feels no need to include any justification for it.

If $X$ is a product of Eilenberg-MacLane spaces then the map $$\eta^*\colon \pi_2(X) = [S^2,X]\to[S^3,X]=\pi_3(X)$$ is easily seen to be zero (where $\eta\colon S^3\to S^2$ is the Hopf map). However, standard calculations give: \begin{align*} \pi_2(QS^0) & =\pi_2^S(S^0)=\mathbb{Z}/2.\eta^2\\ \pi_3(QS^0) &= \pi_3^S(S^0)=\mathbb{Z}/24.\nu \end{align*} with $\eta^3=12\nu$. Thus, $\eta^*$ is nonzero in this context.
• @PeterMay That's true, but the Dyer-Lashof operations depend on the infinite loop structure, which is not unique. On $Q_0S^0$ there is at least the additive structure, and the multiplicative structure translated from $Q_1S^0$, and probably infinitely many others. I don't see a simple argument that there cannot be an exotic infinite loop structure with trivial Dyer-Lashof operations. – Neil Strickland Feb 26 '17 at 22:22
• Equivalent, really, to Neil's answer is that the ring structure of the mod 2 cohomology of $Q_0S^0$ doesn't look like the cohomology of a product of Eilenberg MacLane spaces by the time one gets to dimension 4: there is a nonzero 2 dimensional class whose square is zero. – Nicholas Kuhn Feb 28 '17 at 13:21