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

For the conjugation action, fixed elements have coefficients constant on $G$-conjugacy classes. Therefore $(\mathcal{O}G{)}^G=Z(\mathcal{O}G)$. More generally an $\mathcal{O}$-basis of $(\mathcal{O}G{)}^H$ consists of sums of the $H$-conjugation orbits on $G$.

For $a\in (\mathcal{O}G{)}^H$ define the relative trace, or transfer, byThe expression is independent of the coset representatives and is $G$-fixed. If $z\in Z(\mathcal{O}G)$, then $z{\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_H^G(za)$; hence its image is a transfer ideal of conjugation-fixed elements in the center of an associative algebra.

Let $P$ be a Sylow $p$-subgroup of $H$. Trace transitivity gives ${\mathrm{Tr}}_P^G={\mathrm{Tr}}_H^G{\mathrm{Tr}}_P^H$, proving inclusion of the $P$-image in the $H$-image. Conversely $[H:P]$ is a unit of $\mathcal{O}$, and for an $H$-fixed element $a$,Thus ${\mathrm{Tr}}_H^G(a)={\mathrm{Tr}}_P^G(a/[H:P])$, proving the reverse inclusion and