TITLE: $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
QUESTION [50 upvotes]: Recently, prompted by considerations in conformal field theory, I was lead 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 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 seems 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?

[Added later]:
My paper on WZW models and $H^4(BG,\mathbb Z)$ has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$.

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 it 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 $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$
The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows:
$$
\begin{matrix}
\vdots & \vdots\\
H^4(B\tilde G) & 0 & \vdots & \vdots & & \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots\\
\mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\
\end{matrix}
$$
and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$
cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

REPLY [35 votes]: I try to give an argument without spectral sequences, not sure if this can be considered non-computational though. At least, there is a non-computational syllabus: torsion classes in $H^4(BG,\mathbb{Z})$ would be characteristic classes of torsion bundles over $S^3$ but the latter have to be trivial. 
Now for a slightly more detailed argument: first, the coefficient formula tells us that torsion in $H^4(BG,\mathbb{Z})$ comes from torsion in $H_3(BG,\mathbb{Z})$ because torsion in $H_4(BG,\mathbb{Z})$ would not survive the dualization. Since $G$ is connected, the Hurewicz theorem provides a surjection $\pi_3(BG)\to H_3(BG,\mathbb{Z})$, so any torsion class in $H_3(BG,\mathbb{Z})$ comes from a map $S^3\to BG$ classifying a $G$-bundle over $S^3$. The corresponding clutching map $S^2\to G$ is homotopic to the constant map because $\pi_2(G)$ is trivial. So we find that any $G$-bundle on $S^3$ is trivial, hence $H_3$ and therefore $H^4$ of $BG$ do not have non-trivial torsion.