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 $Spin(n)$.

[OK, a short train ride later, now I'm home from work. To continue...]

The fibre of the 3-connected cover is a 2-type, and in the case of $Spin(n)$ this is a $K(\mathbb{Z},2)$, so at the very least, $String(n)$ can't be finite-dimensional. If one could construct a primitive[1] $PU(\mathcal{H})$-bundle on $Spin(n)$ whose Dixmier-Douady classs was the generator $\langle -,[-,]\rangle \in H^3(Spin(n))$, then you would have an infinite-dimensional Lie group model for $String(G)$ (here $\mathcal{H}$ is a infinite-dimensional separable Hilbert space, $PU(\mathcal{H})$ is then a smooth model for $K(\mathbb{Z},2)$).

([1] Primitive in the sense that for the group operations $G\times G\to G$ and $(-)^{-1}:G\to G$ there are bundle maps covering them.)

I don't know if this is possible or not, but I'm sure this idea has occurred to someone before, and since we haven't seen it, there might be a reason (well, I haven't seen it and everyone goes on about $String$ only being a topological group).

isa Lie group. So either the connected component of the unit is a torus of $\pi_3\neq 0$. I presume the author meant something else but I'm not sure what exactly. – algori Sep 24 '10 at 7:37or$\pi_3\neq 0$. Argh! – algori Sep 24 '10 at 7:45