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Corrected an 'n' an an 'n - 1' and TeXified.
Denis Nardin
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For the classical Lie groups, I think that an easy way to obtain the result is through the fibrations:

$SO(n-1)\to SO(n)\to S^{n-1}$,

$SU(n-1)\to SU(n)\to S^{2n-1}$,

$SP(n-1)\to SP(n)\to S^{4n-1}$

and the homotopy long exact sequence and $\pi_m(S^n) = 0$ for $m$ less than $n$, and $\pi_2(SO(2)) = \pi_2(SU(2))=0$ and the isomorphism of $SP(2)$ and $SO(5)$.