Let $D$ act on the <group algebra> $kG$ by conjugation. Its fixed-point algebra is
$$
(kG)^D=\{a\in kG:dad^{-1}=a\text{ for every }d\in D\},
$$
and the <centralizer> $C_G(D)$ consists of the elements of $G$ commuting with every element of $D$. The <Brauer morphism> is
$$
\beta=\operatorname{Br}_D:(kG)^D\longrightarrow kC_G(D),
\qquad
\sum_{g\in G}a_gg\longmapsto\sum_{g\in C_G(D)}a_gg.
$$
Thus $\beta$ deletes the coefficients of basis elements outside $C_G(D)$. The nonfixed $D$-orbits have cardinality divisible by $p$, so the usual orbit argument shows that this projection is a unital <ring homomorphism>. Hence
$$
\boxed{\beta=\operatorname{Br}_D:(kG)^D\to kC_G(D).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 6 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-138/6/b/i/solution