Let $G$ be a compact, connected Lie group. There is an Atiyah–Hirzebruch spectral sequence 


$$H^\*(BG;K^\*) \implies K^\*(BG)$$


connecting $H^\*BG$, which generally contains torsion, with $K^\*BG \cong \widehat R(G)$, which does not. 


Is it known whether the torsion situation "only improves"? 
More precisely, let $P\_r \subsetneq \mathbb N$ be the set of torsion primes of the $r$th page: 


1. **Is it always the case that $P\_{r+1} \subseteq P\_r$**?
2. **Is it at least true that $P\_r \subseteq P\_2$**?



Torsion in the Atiyah–Hirzebruch spectral sequence of a classifying space
Of course, in any spectral sequence $E\_{r+1}$ is a subquotient of $E\_r$ (the kernel of $d\_r$ divided by the image of $d\_r$). And in general new torsion can appear in the sense of torsion elements in $ker/im$ that are not represented by torsion elements in $ker$. 


But this cannot happen when the spectral sequence is rationally trivial, that is, when the image of $d\_r$ is in the torsion subgroup. So in that case the torsion subgroup of $E\_{r+1}$ is a subquotient of that of $E\_r$. 


This holds for the Atiyah-Hirzebruch spectral sequence of any simply connected space and any cohomology theory $k^\ast$. The reason is that $k^\ast\otimes \mathbb Q$ is a product of ordinary cohomology theories. 


So, yes the torsion situation only improves. 


Edit: I overstated things a bit. Tyler Lawson makes a point that I had not appreciated: at the level of generality of my answer, tensoring the AHSS with $\mathbb Q$ does not yield the AHSS for another cohomology theory. The argument sketched above applies to the homology AHSS of any simply connected space, or spectrum, but in the case of cohomology the conclusion is false in general. Think of the case when $k$ has only two nontrivial coefficient groups $k^0(\ast)=A$ and $k^n(\ast)=B$, $n>0$. The one nontrivial differential $d\_{n+1}$ is a stable cohomology operation $H^\bullet(X;B)\to H^{\bullet +n+1}(X;A)$, corresponding to an element of $H^{n+1}(HB;A)\cong Hom(H\_{n+1}(HB),A)\oplus Ext(H\_{n}(HB),A)$. This group need not be a torsion group, even though $H\_n(HB)$ and $H\_{n+1}(HB)$ are torsion groups for any abelian group $B$. But it's OK if $B$ is finitely generated.


