Brauer pair branching for $C_3⋊D_8$ in characteristic two

The automorphisms of $C_3=⟨a⟩$ assigned to $r,s$ are respectively identity and inversion; they satisfy the relations of $D_8$. Thus the semidirect product exists and has unique normal forms $a^i{r}^j{s}^{ϵ}$, $0\le i<3$, $0\le j<4$, $0\le ϵ<2$. The presentation reduces every word to this form and maps onto that product, so it defines exactly this group of order $24$. In particular $Q⊴P$ and $P\cong D_8$.

Conjugation by $a^i{r}^j{s}^{ϵ}$ sends $r$ to $r^{(-1{)}^{ϵ}}$, so centralizing $Q$ requires $ϵ=0$. Therefore $C_G(Q)=⟨a⟩\times ⟨r⟩$. Among its elements, commuting with $s$ requires $a^i{r}^j=a^{-i}{r}^{-j}$, hence $i=0$ and $j$ even. Thus $C_G(P)=⟨r^2⟩\cong C_2$.

Write $e_m={\sum }_{i=0}^2{\omega }^{-mi}{a}^i$ for $m=0,1,2$. The coefficient of $a^l$ in $e_m{e}_n$ is ${\omega }^{-nl}{\sum }_i{\omega }^{(n-m)i}$. The inner sum is $0$ for $m\mathrm{\ne }n$ by $1+\omega +{\omega }^2=0$, and $3=1$ for $m=n$. Thus $e_m{e}_n={\delta }_{mn}{e}_m$ and ${\sum }_m{e}_m=1$, and $a{e}_m={\omega }^m{e}_m$. Each factor $e_{m}k{C}_G(Q)$ is consequently isomorphic to $k{C}_4=k[u]\mathrm{/}((u-1{)}^4)$. Its elements with nonzero constant coefficient in $u-1$ have a finite geometric inverse; the others are nilpotent. It is local, and an idempotent in it is $0$ or $1$ since one of it and its complement is a unit. Hence these are exactly the three primitive central blocks.

Conjugation by $r$ fixes each $e_m$, while $s$ sends $e_m$ to $e_{-m}$. Thus $e_0$ is the only $P$-stable block. In the relative Brauer projection of $e_0$, its terms $a,a^2$ do not centralize $P$, so only $1$ remains: ${\mathrm{Br}}_{P\mathrm{/}Q}(e_0)=1$. Also $k{C}_G(P)\cong k[v]\mathrm{/}((v-1{)}^2)$ is local by the same unit calculation, with sole block $1$. Therefore $(Q,e_0)⊴(P,1)$. Neither $e_1$ nor $e_2$ is stable, so [F1], applicable since $Q⊴P$, excludes both other inclusions. The fixed sum $e_1+e_2=a+a^2$ has relative image zero, consistently with this conclusion.