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 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 <i>meaning</i>. 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:

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

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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].<br> 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:<br>
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$<br>
from which is follows that for any finite abelian group $A$<br>
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$<br>
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$<br>
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

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 differential $H^4(B\tilde G)\to H^5(K(\pi,2))$
cannot create torsion in degree four. QED