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David Roberts
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The result is that a compact, connected simple Lie group $G$ has $\pi_3(G) = \mathbb{Z}$. Simple covering space or subgroups arguments should get you to $SO(n)$ which is all that matters. For that matter start with the 1-connected $Spin(n)$.

David Roberts
  • 35.4k
  • 11
  • 124
  • 348