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?)



The first element in the stable homotopy of a $K(\mathbb{Z}/2, n)$
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.


