Maximal Brauer pairs exist and are conjugate

There are finitely many $p$-subgroups and finitely many central primitive idempotents in each centralizer algebra. The set of $b$-pairs is nonempty because it contains $(1,b)$, and therefore it has a maximal pair $(P,e)$. Let $z$ be the sum of the distinct $N_G(P)$-conjugates of $e$. They are orthogonal block idempotents of $k{C}_G(P)$, so $z^2=z$, and $z$ is $N_G(P)$-fixed. Every such conjugate pair belongs to $b$ by [F1], so ${\mathrm{Br}}_P(b)z=z$. The pair relation is a partial order by [F8].

We use the elementary normalizer condition for a finite $p$-group $T$: if $P<T$, then $P<N_T(P)$. Indeed let $P$ act on $T\mathrm{/}P$ by left multiplication. Nonfixed orbits have size divisible by $p$ by orbit-stabilizer. The number of fixed cosets is therefore congruent to $[T:P]=0$ modulo $p$. It is positive since $P$ is fixed, and fixed cosets are exactly $N_T(P)\mathrm{/}P$. Hence that quotient has at least $p$ elements.

If $P⊴R$ for a strictly larger $p$-subgroup $R$, partition these summands into $R$-orbits. Non-singletons have zero relative image by [F3]. For an $R$-fixed summand $f$, a nonzero relative image is a central idempotent and hence contains a block $d$ of $k{C}_G(R)$. Centrality follows because $f$ commutes with that algebra; multiplicativity gives idempotence. The stable block and this product satisfy the normal criterion [F8], so $(P,f)⊴(R,d)$, and [F1] makes this a larger $b$-pair. Conjugating back by a normalizer element contradicts maximality of $(P,e)$. Thus ${\mathrm{Br}}_{R\mathrm{/}P}(z)=0$ for every such $R$.

If $x\in C_G(P)$ has nonzero coefficient in $z$, take a Sylow subgroup $T$ of $C_G(x)$ containing $P$, using [F7]. If $T>P$, put $R=N_T(P)>P$ by step 1.2. Then $P⊴R$, and $x\in C_G(R)$, so its nonzero coefficient survives ${\mathrm{Br}}_{R\mathrm{/}P}(z)$, contradicting step 2.1. Thus $P$ is Sylow in $C_G(x)$ for every support element $x$.

Suppose $x$ is in this support and $y=gx{g}^{-1}\in C_G(P)$. The groups $gP{g}^{-1}$ and $P$ are Sylow in $C_G(y)$: the former by step 3.1 and conjugation, the latter by equality of centralizer orders. Choose $c\in C_G(y)$ with $cgP{g}^{-1}{c}^{-1}=P$ by [F7]. Then $cg\in N_G(P)$ and $(cg)x(cg{)}^{-1}=y$. Normalizer invariance of $z$ gives equal coefficients at $x,y$. Consequently each $G$-conjugacy class meeting the support has constant coefficient on its whole intersection with $C_G(P)$; classes not meeting the support have coefficient zero throughout. Summing these coefficients times full $G$-class sums constructs $a\in Z(kG)$ with ${\mathrm{Br}}_P(a)=z$.

The restriction of ${\mathrm{Br}}_P$ to $Z(kG)$ identifies its image with a finite commutative quotient by its kernel, by [F5]. Lift its idempotent $z$ to a central idempotent $c$ by [F4]. Then $bc$ is a central idempotent beneath primitive $b$, hence is $0$ or $b$. Its image is ${\mathrm{Br}}_P(b)z=z\mathrm{\ne }0$, so $bc=b$. Also ${\mathrm{Br}}_P(bc)={\mathrm{Br}}_P(b)$ and equals $z$, proving ${\mathrm{Br}}_P(b)=z$.

If $S>P$ had ${\mathrm{Br}}_S(b)\mathrm{\ne }0$, let $R=N_S(P)>P$. Coefficient projection gives ${\mathrm{Br}}_R(b)={\mathrm{Br}}_{R\mathrm{/}P}({\mathrm{Br}}_P(b))=0$ by steps 2.1 and 5.1. Since $C_G(S)\subseteq C_G(R)$, projecting further would give ${\mathrm{Br}}_S(b)=0$, a contradiction. Thus $P$ is maximal nonzero support, hence a defect group by [F2].

For any other maximal pair $(Q,f)$, step 6.1 makes $Q$ a defect group. By [F2] conjugate it so that $Q=P$. Membership says ${\mathrm{Br}}_P(b)f=f$. Step 5.1 writes this image as exactly the sum of the distinct normalizer conjugates of $e$. Orthogonality of central blocks forces $f$ to be one of these conjugates. This proves the single $G$-conjugacy class, as well as the asserted image formula.