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The first element in the stable homotopy groups of a $K(\mathbb{Z}/2, n)$ (which is outside the range of the Freudenthal suspension theorem) is $\pi_{2n} K(\mathbb{Z}/2, n) \simeq \mathbb{Z}/2$. In particular, there is a unique nontrivial stable map $S^{2n} \to K(\mathbb{Z}/2, n)$. This map is necessarily zero in cohomology.

In what Adams filtration can we detect this map? (Can we detect it using some other cohomology theory?)

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The mod 2 cohomology $H^*(K(\mathbb{Z}/2,n))$ is freely generated over the Steenrod algebra by the canonical class $\iota\in H^n$ in degrees $*\leq 2n$, and the only relation introduced in degree $2n+1$ is $Sq^{n+1}(\iota)=0$. Thus the Adams spectral sequence for $K(\mathbb{Z}/2,n)$ has classes in bidegrees $(0,n)$ (from $\iota$) and $(1,2n+1)$ (from the relation), and nothing else in bidegrees $(s,t)$ with $t-s<2n$. Both of these classes must survive the spectral sequence for degree reasons. This shows that there is a nontrivial stable map $S^{2n}\to K(\mathbb{Z}/2,n)$ of Adams filtration 1.

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