Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 4 / 4 / b

Use the basis of $D$-conjugation orbit sums in $kG$. A singleton orbit is exactly an element of $C_G(D)$, so these basis vectors span the subring $k{C}_G(D)$. For a nonsingleton orbit with representative $x$, its stabilizer is the proper subgroup $E=C_D(x)<D$, and its orbit sum is ${\mathrm{Tr}}_E^D(x)$. Thus every nonsingleton basis vector belongs to the sum of proper-subgroup transfer images.

Conversely the coefficient at a $D$-fixed basis element $x\in C_G(D)$ in ${\mathrm{Tr}}_E^D(a)$ is $[D:E]a_x=0$ in characteristic $p$ whenever $E<D$. Hence the two spans intersect trivially, givingFor $u\in (kG{)}^D$, multiplication by a transfer element satisfies $u{\mathrm{Tr}}_E^D(a)={\mathrm{Tr}}_E^D(ua)$ and ${\mathrm{Tr}}_E^D(a)u={\mathrm{Tr}}_E^D(au)$. Therefore the second summand is a two-sided ideal, as required; it need not be nilpotent in this general setting.