Brauer's Second Main Theorem

Write the ordinary restriction as ${\mathrm{Res}}_H^G\chi ={\sum }_{\zeta \in {\mathrm{Irr}}_K(H)}{n}_{\chi ,\zeta }\zeta \mathrm{.}$ For a block $c$ of $kH$, its lifted idempotent selects exactly the ordinary constituents in $c$, so ${\chi }_c(uv)={\sum }_{\zeta \in {\mathrm{Irr}}_K(H,c)}{n}_{\chi ,\zeta }{\lambda }_{u,\zeta }\zeta (v)\mathrm{.}$ On $p$-regular $v$, expand each $\zeta (v)$ by ordinary decomposition numbers. F4 deletes all terms outside the block $c$. Conversely, in the defining formula for $d_{\chi ,\phi }^u$ with $\phi \in \mathrm{IBr}(H,c)$, F4 deletes every $\zeta$ outside $c$. Therefore ${\chi }_c(uv)={\sum }_{\phi \in \mathrm{IBr}(H,c)}{d}_{\chi ,\phi }^u\phi (v)$ for every $p$-regular $v\in H$.

Suppose $c^G\mathrm{\ne }B$. F3 makes the left side of the last display zero for every $p$-regular $v$. Linear independence in F5 then gives $d_{\chi ,\phi }^u=0$ for every $\phi \in \mathrm{IBr}(H,c)$. The contrapositive is the asserted support implication.

F1 gives the full expansion $\chi (uv)={\sum }_c{\sum }_{\phi \in \mathrm{IBr}(H,c)}{d}_{\chi ,\phi }^u\phi (v),$ where F4 partitions the Brauer basis by its unique blocks. Step 2.1 deletes exactly the summands with $c^G\mathrm{\ne }B$ and proves the displayed restricted formula in the Statement. Conversely, if that restricted formula holds, subtracting it from F1's full expansion and applying F5 block by block forces every coefficient in a noninducing block to be zero, recovering the support implication.

When $u=1$, one has $H=G$, $c^G=c$, and $d_{\chi ,\phi }^1=d_{\chi ,\phi }$; the theorem becomes ordinary block diagonality from F4. Empty local Brauer-character sets contribute empty sums, and no converse asserting that a permitted coefficient is nonzero has been used. Algebraic closedness and AC enter exactly through F2–F3.