We know that there is a map from $h:\pi\_{i}^{st}(pt)\rightarrow KO\_{i}(pt)$ and we know all the $KO\_{i}(pt)$ by Bott periodicy: they are $Z, Z\_{2},Z\_{2},0,Z,0,0,0$. We also know $\pi\_{i}^{st}(pt)$ for i=0~7.


My question is how to determine the image of $h$? 
In particular (which is the case I care most), how to prove that when $i=1$, the Hurewice image of the hopf map is non zero in $KO\_{1}(pt)$? 


I can't find this computation in any reference although this is standard. Can anyone tell me about the reference? 



How to compute the Hurewicz image of a stable map into real K theory
For $i = 1$ you can argue using the fact that $S\_n \to O(n)$ induces an isomorphism on abelianizations. $\pi\_i^{st}$ are the homotopy groups of the group completion of $\coprod\_n BS\_n$, and at least for $i$ positive $KO\_i$ are the homotopy groups of the group completion of $\coprod\_n BO\_n$. 


For $i = 2$, you can argue that the map is surjective using Pontrjagin-Thom construction and the Atiyah invariant. $\pi\_2^{st}$ is the set of cobordism classes of framed surfaces, and the map to $KO\_2 = \mathbf{Z}/2$ factors through the "Atiyah invariant'' of spin surfaces, I'm not sure whether that boils down to something elementary.


For $i = 4,8,12,16,20,\dots$ the map is zero because the domain of the map is torsion.


