Quickest self-contained way of finding $\pi_1(\text{SU}(2))$ and $\pi_1(\text{SO}(3))$?

Question

As the question title suggests, what is the quickest self-contained way of computing $\pi_1(\text{SU}(2))$ and $\pi_1(\text{SO}(3))$?

Answer 1

The quickest way to compute $\pi_1(SU(2))$ by itself is to observe that it is diffeomorphic to $S^3$, which is easily seen to be simply connected.

The quickest way to compute $\pi_1(SO(3))$ by itself is, as Mariano notes, to observe that it acts on $S^2$ with stabilizer $SO(2)$, obtain a fibration and use the long exact sequence in homotopy groups.

If you complain that method is not self-contained, you can instead do the following: Observe that $SU(2)$ double covers $SO(3)$, so $\pi_1(SO(3))$ must be the group of order two. In my opinion this is slightly more work, but can be done with elementary differential topology.