Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups (and in particular, the fact that every connected Lie group admits a finite cover that is a product of a torus with a simply connected group) and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free, seem to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which is follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{n \choose 2},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{n \choose 2} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED