We fix an integer $n$ and consider the stabilization map $O(n)\to O$.
Using rational methods one can easily check that the map
$\pi\_{4i-1}(O(n))\to \pi\_{4i-1}(O)\cong\mathbb{Z}$ vanishes for sufficiently large $i$. Is a similar fact known for the map
$$\pi\_{8i}(O(n))\to \pi\_{8i}(O)\cong\mathbb{Z}/2\mathbb{Z}\ ?$$



Homotopy of orthogonal groups in the unstable range
This is false. In fact, for all sufficiently large n the maps $\pi\_{8i}(O(n)) \to \pi\_{8i}(O) \cong \Bbb Z/2$ are surjective for all $i$.


One possible tool for proving this has to do with $v\_1$-periodic homotopy theory. For sufficiently large $d$, the mod-2 Moore space $W = \Sigma^{d-1} \Bbb{RP}^2$, which is the cofiber of a degree-2 map $S^d \to S^d$, has a certain type of "periodic" self-map $\Sigma^8 W \to W$. Any $2$-torsion element $\alpha \in \pi\_{d+k} X$ extends from a map $S^{d+k} \to X$ to a map $\Sigma^k W \to X$, and we can precompose to get maps $\Sigma^{8+k} W \to X$ which determine a new $2$-torsion element $\alpha' \in \pi\_{d+k+8} X$.


It so happens that, in the case of the orthogonal group $O$, this construction $\alpha \mapsto \alpha'$ gives an isomorphism $\pi\_{8i} O \to \pi\_{8i+8} O$, despite the possible indeterminacy in these definitions. Therefore, if one of these classes lifts to $\pi\_{8i} O(n)$ and $i$ is bigger than $d$, the next one lifts to $\pi\_{8i+8} O(n)$.


