Brauers First Main Theorem

Let $\lambda :Z(kG)\to k$ be a unital $k$-algebra homomorphism. Each block idempotent maps to an idempotent of a field, hence zero or one. Orthogonality prevents two values from being one, and their sum is one by F1, so exactly one value is one, at an idempotent $b$. For any $z$, $\lambda (z)=\lambda (bz)$; thus $\lambda$ factors through the single center $Z(B)$.

Fix a block $B\mathrm{\ne }0$. Among dimensions of proper left ideals of $B$ there is a largest, since zero is proper and dimensions are integers less than ${\dim }_{k}B$. A left ideal of that dimension is maximal, and its quotient $S$ is a nonzero simple $B$-module. Extending the action through $kG\to B$ makes it a simple $kG$-module. Each $z\in Z(B)$ acts as a module endomorphism of $S$, hence as a unique scalar by F3. These scalars define a unital $k$-algebra map $\chi :Z(B)\to k$. It is surjective because scalar multiples of $b$ act by those scalars.

The kernel of $\chi$ is a maximal ideal and therefore is F2's unique maximal ideal $J$. Any other unital $k$-algebra map from $Z(B)$ to $k$ has the same kernel. For $z\in Z(B)$, step 1.2 gives $z-\chi (z)b\in J$, so that map must send $z$ to $\chi (z)$. Thus $\chi$ is independent of $S$ and is the unique such map. Extending by ${\lambda }_b(z)=\chi (bz)$ and using step 1.1 proves the bijection and the stated characterization. The group algebra is nonzero, so its block set is not empty; a sole block gives a sole map. Selecting one finite-dimensional ideal for an existence proof uses no AC.

Choose a defect group $E$ of $B$ by F2 and F4. Relative $\mathrm{\Delta }E$-projectivity writes $B$ as a summand of a module induced from $\mathrm{\Delta }E$. Restrict it to $H\times H$ and apply F3. By F1 and finite summand extraction, $b$ is relatively $(H\times H)\cap x\mathrm{\Delta }E{x}^{-1}$-projective for some $x\in G\times G$. Since $\mathrm{\Delta }D$ is a vertex by F4, F3 places it in an $(H\times H)$-conjugate of that intersection, hence in a $(G\times G)$-conjugate of $\mathrm{\Delta }E$.

Write that conjugating element as $(g_1,g_2)$. Projecting $\mathrm{\Delta }D\le (g_1,g_2)\mathrm{\Delta }E(g_1,g_2{)}^{-1}$ onto the first coordinate gives $D\le g_{1}E{g}_1^{-1}$. Conjugating a diagonal vertex simultaneously by $(g_1,g_1)$ shows this conjugate of $E$ is again a defect group of $B$. This proves the required containment. If $D=1$ the conclusion is automatic; if $H=G$, F1 gives $B=b$ and equality is possible. Neither argument infers equality in general. All selections involve finite subgroup sets and finite decompositions, with no additional AC.

By F2, as an $H\times H$-module $M=kG$ splits over its double-coset orbits into $kH\oplus {⨁}_{t\mathrm{\notin }H}k[HtH]$. Each orbit module is induced from its point stabilizer: the map from stabilizer cosets to orbit points is a bijection, exactly as in F2. If an indecomposable summand of $k[HtH]$ had a vertex containing an $(H\times H)$-conjugate of $\mathrm{\Delta }D$, F3 would place that diagonal conjugate in a conjugate of the point stabilizer. Equivalently $\mathrm{\Delta }D$ would fix some point $t^{\mathrm{\prime }}\in HtH$, so $d{t}^{\mathrm{\prime }}=t^{\mathrm{\prime }}d$ for every $d\in D$. This puts $t^{\mathrm{\prime }}\in C_G(D)\le H$, impossible when $t\mathrm{\notin }H$. The same argument proves the exclusion for any $p$-subgroup $Q\le H$ with $C_G(Q)\le H$, replacing $D$ by $Q$.

The identity orbit $kH$ contains $b$ exactly once in its block decomposition, by F1. By F4, F9 and step 1.1 no other orbit contains an isomorphic summand. Thus $b$ has multiplicity one in $M$. Decomposing $M=kG$ into global blocks and applying F8 shows exactly one global block $B$ has $b$ as a restriction summand. By F1 this is $b^G$. This proves existence without the extra normalizer assumption.

We identify its central character carefully. Use the decomposition $M=b\oplus X$, where the projection onto $b$ is $r(a)=e{\pi }_H(a)$ and ${\pi }_H$ deletes coefficients outside $H$. Decompose $X$ into indecomposables $X_j$ by F8; none is isomorphic to $b$ by step 2.1. Any composite $b\to X_j\to b$ is a nonunit in ${\mathrm{End}}_{H\times H}(b)$: if invertible it would split $b$ from $X_j$, forcing an isomorphism by indecomposability. This endomorphism ring is $Z(b)$, as in F6's block-center construction; its nonunits form the nilpotent ideal $J_b$. Therefore for $T\in {\mathrm{End}}_{H\times H}(M)$ the function $\chi (T)={\lambda }_b(rT{\mathrm{\mid }}_b)$ is a unital algebra homomorphism: in the corner of a composite, all cross-composites through the $X_j$ vanish modulo $J_b$, leaving the product of the two corners.

Let $E_C$ be projection of $M$ onto a global block $C$. If $C\mathrm{\ne }B$, decompose its restriction into indecomposables, none isomorphic to $b$, and step 3.1 gives $\chi (E_C)=0$. Since the projections sum to the identity, $\chi (E_B)=1$. For $z\in Z(kG)$ let $L_z$ be multiplication by $z$. On $B$, multiplication by $z-{\lambda }_B(z)$ is nilpotent by F6. Hence $(L_z-{\lambda }_B(z))E_B$ is nilpotent on $M$, and applying the field-valued homomorphism $\chi$ gives $\chi (L_z)={\lambda }_B(z)$. But its corner on the explicit $b\subset kH$ is multiplication by $e{\pi }_H(z)$, because ${\pi }_H$ is an $H$-bimodule projection. Thus ${\lambda }_B(z)={\lambda }_b({\pi }_H(z))\mathrm{.}$ Centrality of $z$ makes ${\pi }_H(z)\in Z(kH)$.

Now assume $H\le N_G(D)$, so $D⊴H$. Let $S$ be a simple $b$-module, whose existence and scalar character are proved in F6. F7 says every $d\in D$ acts trivially on $S$. Expand ${\pi }_H(z)$ in group elements and partition $H\setminus C_G(D)$ into conjugation orbits of $D$. Each orbit has size a power of $p$ greater than one; its coefficients in $z$ are constant. All conjugate elements have the same operator on $S$, so its orbit sum acts as zero in characteristic $p$. The remaining terms are precisely ${\mathrm{Br}}_D(z)$ by F5. Since $H$ normalizes $D$, it preserves $C_G(D)$ and this projection is central in $kH$. Consequently the two central elements have the same scalar on $S$, giving ${\lambda }_b({\pi }_H(z))={\lambda }_b({\mathrm{Br}}_D(z))$. Combine with step 4.1 to prove the formula.

For $D=1$, the containment assumption forces $H=G$ and all projections in the formula are identity. For $H=G$ induction is already the identity by F1. Empty off-identity double-coset families and empty noncentralizing orbit families simply contribute zero in the above sums. Every decomposition and orbit calculation is finite; no AC is added. The extra normalizer assumption was used only in step 5.1, so the formula has not been asserted outside its stated domain.

Put $b=c^H$ and $B=b^G$. By F1 there are split inclusions and retractions $i:c\to {\mathrm{Res}}_{K\times K}b$, $r$ back, and $j:b\to {\mathrm{Res}}_{H\times H}B$, $s$ back, with $ri=1_c$ and $sj=1_b$. Restricting $j,s$ to $K\times K$ retains their composite identity. Therefore $(\mathrm{Res}j)i$ and $r(\mathrm{Res}s)$ have composite $1_c$, exhibiting $c$ as a summand of ${\mathrm{Res}}_{K\times K}B$.

Since $c^G$ is defined, F1 makes it the unique global block with this summand property. Step 1.1 proves that $B$ has that property, so $c^G=B=(c^H{)}^G$. The same map composition works when two or all three groups coincide. The proof assumes the existence of all three blocks and does not deduce the third definedness from the first two. It composes finitely many given maps and uses no AC.

For $z\in Z(kH)$, normality of $D$ makes $C_H(D)$ stable under $H$, so ${\mathrm{Br}}_D(z)$ is central in $kH$. On a simple module, each $D$-conjugation orbit outside $C_H(D)$ consists of elements with the same action operator by F2. Its coefficients in $z$ are equal, and its length is a positive power of $p$ greater than one. Its sum therefore acts as zero. Removing all these orbits leaves F3's projection. Hence every block character satisfies ${\lambda }_b(z)={\lambda }_b({\mathrm{Br}}_D(z))$, by its simple-module construction in F1.

If $e$ is central idempotent, its image is a central idempotent by F4 and step 1.1. A central idempotent is the sum of a subset of primitive block idempotents: multiply it by each primitive block and use primitivity to obtain either that block idempotent or zero. F1's characters read exactly the indicator of this subset. Step 1.1 says the two indicators for $e$ and its image agree, so the idempotents themselves agree. This includes $e=0,1$, $D=1$, and an empty set of removed orbits. All arguments involve finite sums and require no AC.

Take a maximal $B$-pair $(P,a)$ from F2. By F3 its subgroup $P$ is a defect group. Apply F4 to $P$ and $D$ in both directions: their orders agree, and $P$ is conjugate to $D$. Conjugate the pair to have subgroup literally $D$. Then F2 gives $e={\mathrm{Br}}_D(f)$ as the sum of one $N$-orbit of primitive central idempotents of $k{C}_G(D)$. It is nonzero, idempotent and $N$-invariant, hence central in $kN$.

Suppose $u$ is a central idempotent of $kN$ beneath $e$. Since $D⊴N$, F1 gives $u={\mathrm{Br}}_D(u)\in k{C}_G(D)$. It is central in that algebra, since $C_G(D)\le N$. Thus $u$ is a sum of a subset of the primitive idempotents in step 1.1. Centrality in $kN$ makes this subset $N$-stable, and one transitive orbit has only the empty and whole stable subsets. Hence $u=0$ or $u=e$. So $e$ is primitive in $Z(kN)$ and defines a block $b$. Moreover ${\mathrm{Br}}_D(e)=e\mathrm{\ne }0$. If $S>D$ were a $p$-subgroup of $N$ with ${\mathrm{Br}}_S(e)\mathrm{\ne }0$, F7 and $C_G(S)\subseteq C_G(D)$ would give ${\mathrm{Br}}_S(f)={\mathrm{Br}}_S(e)\mathrm{\ne }0$, contradicting F4 for $f$. Therefore F4, now in $N$, proves $D$ is a defect group of $b$.

Since $C_G(D)\le N$, F5 defines $b^G$ and gives ${\lambda }_{b^G}(f)={\lambda }_b(e)=1$. F6 therefore identifies $b^G=B$. Conversely, if a block $c$ of $kN$ with defect group $D$ induces to $B$, F5 gives $1={\lambda }_B(f)={\lambda }_c(e)$. Since $e$ is primitive by step 2.1, F6 forces $c=b$. This proves uniqueness in the stated defect-$D$ domain as well as the promised existence. For $D=1$, $N=G$ and the construction gives $e=f$; the same holds whenever $N=G$ by F1. All chosen pairs, subgroups and idempotent subsets lie in finite sets; no AC or stronger restriction-summand theorem was used.

We first record the elementary normalizer condition: if $P<T$ are finite $p$-groups, let $P$ act on $T\mathrm{/}P$ by left multiplication. Nonfixed orbit sizes are divisible by $p$, and $[T:P]$ is divisible by $p$. The fixed points are $N_T(P)\mathrm{/}P$ and include the identity coset, so their positive cardinality is divisible by $p$; hence $N_T(P)>P$. Now if $x$ has nonzero coefficient in $e$, F1 puts $x\in C_G(D)$. By F7 take a Sylow subgroup $T$ of $C_G(x)$ containing $D$. If $T>D$, its subgroup $R=N_T(D)>D$ lies in $N$ and centralizes $x$. The coefficient of $x$ survives ${\mathrm{Br}}_R(e)$ by F6, contradicting F3 for the local block. Thus $D$ is Sylow in $C_G(x)$ for every support element $x$ of $e$.

Suppose $y=gx{g}^{-1}$ also centralizes $D$, with $x$ in that support. Both $gD{g}^{-1}$ and $D$ are Sylow in $C_G(y)$: the first by step 1.1, and the second by equal order and $D\le C_G(y)$. F7 supplies $c\in C_G(y)$ with $cgD{g}^{-1}{c}^{-1}=D$. Then $cg\in N$ and $(cg)x(cg{)}^{-1}=y$. Since $e$ is central in $kN$, its coefficients at $x$ and $y$ agree. Therefore its coefficients are constant on every intersection of a $G$-conjugacy class with $C_G(D)$, with zero throughout intersections missing the support. Give each full $G$-class that common coefficient, zero for a class disjoint from $C_G(D)$. This finite class sum is $a\in Z(kG)$ with ${\mathrm{Br}}_D(a)=e$ by F6.

By F5 the image of $Z(kG)$ is a finite commutative quotient algebra. Its idempotent $e$, present by step 2.1, lifts by F4 to a central idempotent $u\in kG$. Write $u$ as a sum of distinct global primitive block idempotents $f_j$ using F8. Their Brauer images are orthogonal idempotents, central in $kN$ because $N$ normalizes $D$, and sum to $e$. Since $e$ is primitive in $Z(kN)$, exactly one image equals $e$ and all others are zero. Let $f$ be that block idempotent and $B=kGf$. F2 applies because $C_G(D)\le N$ and gives ${\lambda }_{b^G}(f)={\lambda }_b(e)=1$. F8 forces $b^G=B$.

Its Brauer image at $D$ is nonzero. If $S>D$ had ${\mathrm{Br}}_S(f)\mathrm{\ne }0$, put $R=N_S(D)>D$ by step 1.1. Then $D\le R\le N$. Since $C_G(S)\subseteq C_G(R)\subseteq C_G(D)$, F6 gives ${\mathrm{Br}}_R(e)={\mathrm{Br}}_R(f)\mathrm{\ne }0$: a nonzero coefficient retained at $S$ is still retained at $R$. This contradicts F3 for the local block $e$. Thus no such $S$ exists, and F3 proves $D$ is a global defect group of $B$. For $D=1$ the normalizer is $G$, and the proof gives the original defect-zero block; when no local defect-$D$ block exists the universal assertion has no inputs. All lifting and subgroup selections here concern finite sets, so no AC or stronger Green restriction theorem is used.

F1 applies to $b$ and gives that the common block $B$ has defect group $D$. Therefore F2 applies to this actual global block and subgroup, giving one primitive local idempotent $e={\mathrm{Br}}_D(f)$, where $f$ is the idempotent of $B$.

Both induced characters take value one at $f$. By F3, ${\lambda }_b(e)={\lambda }_c(e)=1$. Since $e$ is primitive by step 1.1, F4 forces both local blocks to be $kNe$. Hence $b=c$. This proves injectivity even if the set of such blocks has zero or one element. For $D=1$ or $N=G$ it is the identity case of F2. Only finite idempotent identifications occur; no claim about the vertex of an arbitrary restriction summand is used.

F3 makes the displayed assignment a function with the stated codomain. F2 makes it injective, and F1 makes it onto. These are precisely the two bijection conditions, including if either set is empty: F1 and F3 then force the other to be empty as well.

For $g\in G$, conjugation carries $N_G(D)$ to $N_G(gD{g}^{-1})$, sends a block ideal to the conjugate ideal, and transports the double-group action and every split inclusion/retraction by the algebra isomorphism $x\mapsto gx{g}^{-1}$. A diagonal vertex $\mathrm{\Delta }D$ is carried to $\mathrm{\Delta }(gD{g}^{-1})$, since relative induction splittings and subgroup minimality are transported in both directions. Thus conjugating the summand condition defining $b^G$ gives $({}^{g}b{)}^G={}^g(b^G)$ by uniqueness. If $N=G$, restriction to the same double group is identity and each block is its own unique inducing block, so the bijection is identity. In particular $D=1$ gives the identity on defect-zero blocks. This proves equivariance and all boundaries; no preferred representative of a conjugacy class is chosen. F1–F3 are choice-free and the present maps are explicit, so no AC is required.

If $(x,y)$ normalizes $\mathrm{\Delta }D$, conjugation and coordinate projection give $xD{x}^{-1}=yD{y}^{-1}=D$. Thus $x,y\in N$ and $N_{G\times G}(\mathrm{\Delta }D)\le N\times N$. By F1, F2 and F4 both blocks are nonzero indecomposable modules with vertex exactly $\mathrm{\Delta }D$, and the defining induction property gives $b\mid {\mathrm{Res}}_{N\times N}B$. The global vertex is already proved by the Brauer bijection; it is not inferred from this summand relation.

Apply F3 under A1 using the normalizer containment in step 1.1. Its unique vertex-$\mathrm{\Delta }D$ restriction summand must be $b$, and the inverse correspondence sends $b$ to $B$. These are exactly the two Green-correspondence assertions. When $D=1$ or $N=G$, the double-group normalizer interval is the identity case and the correspondence fixes the block. The only added AC use is application of F3, whose inherited finite-length argument declares it; the finite group calculation in step 1.1 and the block bijection do not use AC.

By F1 and F2 take a finite-dimensional $kR$-module $W$ and split $X$ from ${\mathrm{Ind}}_R^{H}W$; one may take $W={\mathrm{Res}}_R^{H}X$ using F2's counit splitting. Tensor the inclusion and retraction with $1_Y$. Their composite remains $1_{X\otimes Y}$, and both are $H$-linear for the diagonal actions. Thus it suffices to identify the tensor of the inducing module.

Define $\mathrm{\Phi }((h\otimes w)\otimes y)=h\otimes (w\otimes h^{-1}y),\mathrm{\Psi }(h\otimes (w\otimes y))=(h\otimes w)\otimes hy\mathrm{.}$ These map between $(kH{\otimes }_{kR}W){\otimes }_{k}Y$ and $kH{\otimes }_{kR}(W{\otimes }_k{\mathrm{Res}}_R^{H}Y)$. The balancing relation $hr\otimes w=h\otimes rw$ is respected by $\mathrm{\Phi }$: moving $r$ from the first tensor factor on its right acts diagonally on $w\otimes r^{-1}{h}^{-1}y$, giving $rw\otimes h^{-1}y$. The same identity verifies $\mathrm{\Psi }$ is balanced. Applying $a\in H$ on the left replaces $h,y$ by $ah,ay$, leaving $h^{-1}y$ unchanged, so $\mathrm{\Phi }$ is $H$-linear. The displayed formulas compose to the identity in both orders. Therefore the tensor in step 1.1 is a summand of an $R$-induced module, proving relative $R$-projectivity.

For the consequence, decompose $X={⨁}_j{X}_j$ using F3 and choose a vertex $R_j$ for each nonzero indecomposable by F4. Step 2.1 makes $X_j\otimes Y$ relatively $R_j$-projective. Any indecomposable summand $Z$ of their finite sum is a summand of one term by F2. F2 then puts a vertex $T$ of $Z$ inside an $H$-conjugate of $R_j$. If $T$ contained a conjugate of $Q$, so would that conjugate of $R_j$, contradicting the hypothesis. This proves the consequence. If either tensor factor is zero the sum has no indecomposable summands. For $R=H$ the assertion is automatic, and for $Q=1$ its hypothesis forces $X=0$. All decompositions, bases and subgroup choices here are finite.

By F6 choose a defect group $D_b$ of $b$ containing the literal $Q$, after conjugating inside $H$. Then $C_G(D_b)\le C_G(Q)\le H$, so F2 defines $b^G$. Suppose for contradiction $b^G\mathrm{\ne }B$. F1 implies $b$ is not a summand of ${\mathrm{Res}}_{H\times H}B$.

Set $M={⨁}_{HtH\mathrm{\ne }H}k[HtH]$. The double-coset decomposition gives $kG=kH\oplus M$ as $H\times H$-modules under F7's double action, and left multiplication by $e$ gives $ekG=b\oplus eM$. This multiplication is an $H\times H$-linear idempotent because $e$ is central in $kH$. Similarly $eB$ is a direct summand of both $B$ restricted and $ekG$. The first conclusion in step 1.1 excludes $b$ from $eB$. By F5 every indecomposable summand of $eB$ must therefore come from $eM$, so $eB$ is a summand of $eM$, hence of $M$.

Restrict $eB$ to $\mathrm{\Delta }H$. For each of its indecomposable summands with vertex $T$, F3's Mackey decomposition puts every vertex of a restricted indecomposable inside a $\mathrm{\Delta }H$-conjugate of $\mathrm{\Delta }H\cap xT{x}^{-1}$ for some $x\in H\times H$. If such a vertex contained a conjugate of $\mathrm{\Delta }Q$, then $T$ would contain an $(H\times H)$-conjugate of $\mathrm{\Delta }Q$, contrary to step 2.1. Identifying $H$ with $\mathrm{\Delta }H$ gives the conjugation action $h:a\mapsto ha{h}^{-1}$ on $eB$. Thus this $kH$-module has no summand whose vertex contains an $H$-conjugate of $Q$. F4 gives the same exclusion for $eB{\otimes }_{k}eV$ with diagonal action.

Define $i:eV\to eB\otimes eV$ by $i(v)=ef\otimes v$, and $r:eB\otimes eV\to eV$ by $r(a\otimes v)=av$. The first is $H$-linear because $h(ef)h^{-1}=ef$; the second is $H$-linear because $(ha{h}^{-1})(hv)=h(av)$. Its image lies in $eV$ since $ea=a$. Since $fV=V$ and $ev=v$ on $eV$, their composite is $ri(v)=efv=v$. Hence $eV$ is a summand of this tensor. Also $eU=U$ because $U$ belongs to $b$, so the given splitting of $U$ from the restriction of $V$, after applying $e$, splits $U$ from $eV$. Consequently $U$ is a summand of $eB\otimes eV$.

This contradicts step 3.1, since $U$ has vertex $Q$. Therefore $b^G=B$. If $Q=1$, the centralizer condition forces $H=G$ and the conclusion is the identity block assignment. If $H=G$ directly, the same conclusion holds. The nonzero module $U$ ensures the splittings in step 4.1 cannot be vacuous. All tensor maps and decompositions are finite and require no AC.

Take $b$ from F1. It is nonzero, so a proper left ideal of largest dimension in $b$ gives a nonzero simple quotient $S$. Extending by zero on the other blocks makes it a simple $kN$-module. By F6, $D=O_p(N)$, in particular $D⊴N$, and F5 makes $D$ act trivially on $S$. Thus $S$ is a simple module of the finite-dimensional quotient group algebra $k[N\mathrm{/}D]$.

Take its projective cover $P\twoheadrightarrow S$ over $k[N\mathrm{/}D]$ using F3; it is finite dimensional, nonzero and indecomposable by F4. Inflate to $N$. The action factors through the quotient, so its submodules and endomorphisms are unchanged and it stays indecomposable. The central idempotents in F9 decompose $P$ into block pieces. The piece for $b$ maps onto $S$, since the idempotent of $b$ acts as identity on $S$. It is therefore nonzero, and indecomposability forces that piece to be all of $P$. Thus $P$ lies in $b$.

F3 realizes $P$ as a summand of a finite free $k[N\mathrm{/}D]$-module. Inflating $k[N\mathrm{/}D]$ gives the permutation module ${\mathrm{Ind}}_D^{N}k$: the map $n\otimes 1\mapsto nD$ identifies their bases and actions. Hence $P$ is relatively $D$-projective. By F11 and F12 it has a vertex $R\le D$, after conjugating in $N$ (normality of $D$ retains this containment). On restriction to $D$, $P$ is a nonzero direct sum of trivial one-dimensional modules, since all of $D$ acts trivially. The trivial $kD$-module has vertex $D$: for $E<D$, every scalar endomorphism has relative trace $[D:E]a=0$ in characteristic $p$, whereas the identity is nonzero, so F10 excludes relative $E$-projectivity. F8 applied to $R\le D$ says the restriction of $P$ has a summand with vertex $R$. All its indecomposable summands are trivial and have vertex $D$, forcing $R=D$.

Apply F7 under A1 to $P$ for $N=N_G(D)$. Its inverse correspondent $V$ is a nonzero indecomposable $kG$-module with vertex $D$, and $P\mid {\mathrm{Res}}_N^{G}V$. F2 applies with $Q=D$, since $D{C}_G(D)\le N$, and identifies the block of $V$ as $b^G=B$. This is the desired module. If $D=1$, the correspondence is identity and the cover already provides the module in $B$. The nonzero simple quotient guarantees no zero object enters. The finite ideal, cover and trace calculations are choice-free; AC is inherited only from F7's declared finite-length chain and selection argument.