Let $s \geq 0$ be fixed. The $J$-homomorphism includes $\pi\_{8s+1}(SO) = \mathbb Z/2$ in $\pi\_{8s+1}^s$, the $(8s+1)$-th stable homotopy group of spheres. 


Now regard $\pi\_{8s+1}^s = \pi\_{8s+1} ((B\Sigma\_{\infty})^+)$ , where $\Sigma\_{\infty}$ is the group of compactly supported permutations of $\mathbb N$. 


The obvious map (i.e. write permutations as permutation matrices) $\Sigma\_{\infty} \to O$ induces a map $\pi\_{8s+1}^s \to \pi\_{8s+1}(BO) = \pi\_{8s}(O)= \mathbb Z/2$. 


Is the composition $\mathbb Z/2 = \pi\_{8s+1} (SO) \to \pi\_{8s+1}^s \to \pi\_{8s} (O) = \mathbb Z/2$ trivial or nontrivial? 


For $s = 0$ this map is nontrivial, since the two elements of $\pi\_1^s$ correspond to odd and even permutations in $\Sigma\_{\infty}$, whereas the two elements in $\pi\_1(BO) = \pi\_0(O)$ correspond to determinant $\pm1$ matrices and the map $\Sigma\_{\infty} \to O$ preserves this structure.


So what happens for positive $s$? [I have a strong feeling that the map is then trivial, but maybe I am wrong...]


Remark: This map factors through $K\_{8s+1}(\mathbb Z)$.



Map between homotopy groups of O, related to J-homomorphism and K-theory of Z
The composition is trivial. Composition with $\eta$ acts trivially on the source and nontrivially on the target, for positive $s$, which implies the claim. This argument does not apply for $s=0$, because $J : SO \to QS^0$ does not deloop.


