Do all Brauer relations for finite groups vanish under the augmentation map?

Question

The Burnside ring $B(G)$ of a finite group $G$ is the ring of all finite $G$-sets under disjoint union and Cartesian product. It is well known that $\mathbb{Q} \otimes B(G)$ has as basis $\{[G/H] \mid H \le_G G \}$ ($\le_G$ means $H$ is taken up to $G$-conjugacy) and there is a natural surjection $\theta$ onto the representation ring $\mathbb{Q} \otimes R(G)$ which has basis the images of $\{[G/H] \mid H \le_G G, \mbox{ $H$ cyclic } \}$. The kernel $ker(\theta)$ is called the set $K(G)$ of Brauer relations. For example, if $G=S_3$, $K(G)$ is generated by a single relation $$[G/\langle () \rangle]-2[G/\langle (1,2) \rangle]-[G/\langle (1,2,3) \rangle]+2[G/G].$$

Observe that applying the augmentation map $\epsilon: [G/H] \mapsto 1$ to this relation gives $1-2-1+2=0$. More generally I have checked that all relations of all groups $G$ with $|G| \le 63$ vanish under $\epsilon$. Is this true in general?

Answer 1

Yes, because $\epsilon$ is equal to the composition of $\theta$ with the map $\mathbb Q\otimes R(G)\to \mathbb Q$ sending the trivial representation to $1$ and all nontrivial irreducible representations to $0$. Since Brauer relations are sent to zero by $\theta$ they are sent to $0$ by $\epsilon$.

To check this, we just need to observe that the representation associated $[G/H]$ is the sum of the one-dimensional trivial representation with some nontrivial irreducible representations. By semismiplicity, we need to check that the $G$-invariants of this representation are one-dimensional, which is clear as any $G$-invariant function on $G/H$ must be constant.

Answer 2

Put another way, (implicitly using Frobenius reciprocity), the augmentation map as you define it counts the multiplicity of the trivial character in the virtual permutation character naturally associated to a $\mathbb{Z}$-linear combination of transitive $G$-sets. On the other hand, if that combination of $G$-sets is sent to zero in the representation ring, its (virtual) character is clearly the zero character, so the trivial character obviously occurs with multiplicity zero. ( Well, I am working integrally here, but the argument adapts to $\mathbb{Q}$-linear combinations).