Block centre locality and trace ideal sums

For $c\in Z(B)$, choose $n$ so the kernels and images of multiplication by $c^n$ stabilize. Their intersection is zero: $c^{n}x$ in the kernel gives $c^{2n}x=0$, hence $c^{n}x=0$. Rank-nullity now gives $B=\ker c^n\oplus \mathrm{im}{c}^n$ as bimodules. Indecomposability forces one summand to be zero. Thus $c$ is nilpotent or multiplication by $c$ is bijective. In the latter case a preimage of $b$ is its inverse and commutes with $B$, so $c$ is a unit in $Z(B)$.

Nonunits in this commutative algebra are nilpotent. A sum of two nilpotents is nilpotent by expanding $(c+d{)}^{r+s}$ when $c^r=d^s=0$; scalar multiplication by any central element also preserves nilpotence. Nonunits therefore form a proper ideal containing all proper ideals, which is the unique maximal ideal. If $b=\sum z_j$ with $z_j$ in specified ideals, some $z_j$ must be a unit, since a sum of nonunits cannot equal $b$. Its ideal contains $z_j^{-1}{z}_j=b$.

For $a\in B^H$ the trace is $G$-fixed, hence central. For $c\in Z(B)$, $c{\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_H^G(ca)$ and $ca\in B^H$. The trace image is a linear subspace stable under central multiplication, hence an ideal, proving the stated application.