What is the third homotopy group of $SO(5)$? I am trying to compute the $\pi_3(SO(5))$, but not getting any idea to compute it. 
As in the case of first and second homotopy groups of $SO(5)$, which can easily be computed using the long exact sequence of homotopy groups. I know that $S^3$ covers $SO(3)$, and $S^3\times S^3$ covers $SO(4)$, but I don’t know whether we have any such covering for $SO(5)$ or not. Any help would be great.
 A: First note that $SO(n)$ has double cover $\operatorname{Spin}(n)$, so for $i > 1$, we have $\pi_i(SO(n)) \cong \pi_i(\operatorname{Spin}(n))$. For small values of $n$, there are exceptional isomorphisms:
\begin{align*}
\operatorname{Spin}(1) &\cong \mathbb{Z}_2\\
\operatorname{Spin}(2) &\cong U(1)\\
\operatorname{Spin}(3) &\cong SU(2) \cong Sp(1)\\
\operatorname{Spin}(4) &\cong SU(2)\times SU(2) \cong Sp(1)\times Sp(1)\\
\operatorname{Spin}(5) &\cong Sp(2)\\
\operatorname{Spin}(6) &\cong SU(4).
\end{align*}
Therefore $\pi_3(SO(5)) \cong \pi_3(\operatorname{Spin}(5)) \cong \pi_3(Sp(2))$.
The long exact sequence in homotopy applied to the fibration $Sp(1) \to Sp(2) \to S^7$ shows that $\pi_3(Sp(2)) \cong \pi_3(Sp(1))$. As $Sp(1)$ is diffeomorphic to $S^3$, we see that $\pi_3(Sp(2)) \cong \mathbb{Z}$ and hence $\pi_3(SO(5)) \cong \mathbb{Z}$.
More generally, $\pi_3(G) \cong \mathbb{Z}$ for any simple Lie group $G$. This was shown by Bott in An application of the Morse theory to the topology of Lie-groups.
