Is there a homomorphism between $\pi_8(S^5)$ and $\pi_8(SO(6))$? I am currently thinking about a physics model related to framed bordism $\Omega_3^{fr}=\mathbb{Z}/24=\pi^s_3$, and the first stable example is $\pi_8(S^5)$, so I was curious about the generator, and happened to see $\pi_8(SO(6))=\mathbb{Z}/24$ as well.
So my question is: is there a homomorphism sending $\pi_8(SO(6))$ to $\pi_8(S^5)$? More generally, what's the relation between homotopy groups $\pi_n(S^m)$ and $\pi_n(SO(m+1))$ in general?
 A: There is a fibration $p : SO(m+1) \to S^m$ with fibre $SO(m)$ which induces a long exact sequence in homotopy
$$\dots \to \pi_n(SO(m)) \to \pi_n(SO(m+1)) \xrightarrow{p_*} \pi_n(S^m) \to \pi_{n-1}(SO(m)) \to \dots$$
For your particular question, we have the fibration $p : SO(6) \to S^5$ with fibre $SO(5)$ which gives 
$$\dots \to \pi_8(SO(5)) \to \pi_8(SO(6)) \to \pi_8(S^5) \to \pi_7(SO(5)) \to \pi_7(SO(6)) \to \pi_7(S^5) \to \pi_6(SO(5)) \to \dots$$
I don't know how to compute these groups myself, but they have been computed by others. From the bottom of page $3$ of this we see that $\pi_8(SO(5)) = 0$, $\pi_7(SO(5)) = \mathbb{Z}$, $\pi_7(SO(6)) = \mathbb{Z}$, and $\pi_6(SO(5)) = 0$. From here we see that $\pi_7(S^5) = \mathbb{Z}_2$ so we have
$$\dots \to 0 \to \pi_8(SO(6)) \xrightarrow{p_*} \pi_8(S^5) \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0 \to \dots$$
As the map $\mathbb{Z} \to \mathbb{Z}_2$ is surjective, it has kernel $2\mathbb{Z}$. So the map $\mathbb{Z} \to \mathbb{Z}$ must be multiplication by $2$ (or $-2$) which is injective. Therefore the map $\pi_8(S^5) \to \mathbb{Z}$ is the zero map, so $p_*$ is an isomorphism.
The fibration always gives rise to a homomorphism $p_* : \pi_n(SO(m+1)) \to \pi_n(S^m)$ but it is not necessarily an isomorphism as $p_* : \pi_7(SO(6)) \to \pi_7(S^5)$ demonstrates.
