 Solution

Put $N=N_G(D)$. For an algebra on which a group acts by conjugation, writefor its transfer ideal of conjugation-fixed elements. In the diagram, $\tau ={\mathrm{Tr}}_D^G$, ${\tau }^{\prime }={\mathrm{Tr}}_D^N$, the upper map is $\beta ={\mathrm{Br}}_D$, and ${\beta }^{\prime }$ is the restriction of ${\mathrm{Br}}_D$ from $(kG{)}^G$ to $(kG{)}_D^G$. Since $N$ normalizes both $D$ and $C_G(D)$, the lower map lands in $(k{C}_G(D){)}_D^N$.

For $a\in (kG{)}^D$, let $D$ act on the left cosets $G/D$. A coset $gD$ is fixed precisely when $g^{-1}Dg\le D$, hence, because the two groups have the same order, precisely when $g\in N$. Every nonfixed orbit has size divisible by $p$. Moreover, after applying ${\mathrm{Br}}_D$, all summands indexed by one $D$-orbit are equal: conjugation by an element of $D$ acts trivially on $k{C}_G(D)$. Those orbits therefore contribute zero in characteristic $p$, while the fixed cosets contribute the trace over $N/D$. ConsequentlyThis is the Brauer morphism and relative trace identity, so the diagram commutes.