Brauers Second Main Theorem

By F2 choose $r,s\in \mathbb{Z}$ with $r{p}^a+sm=1$, and set $g_p=g^{sm},g_{p^{\mathrm{\prime }}}=g^{r{p}^a}\mathrm{.}$ These elements commute because they are powers of $g$, and their product is $g^{sm+r{p}^a}=g$. Also $g_p^{p^a}=1$ and $g_{p^{\mathrm{\prime }}}^m=1$, so their orders divide $p^a$ and $m$, respectively. Thus they have the required order types.

Suppose $g=uv$ is another commuting factorization, with $\mathrm{\mid }u\mathrm{\mid }=p^b$ and $\mathrm{\mid }v\mathrm{\mid }=d$ prime to $p$. Bezout applied to $p^b$ and $d$ gives an exponent $e$ satisfying $e\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{p}^b)$ and $e\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}d)$; hence $g^e=(uv{)}^e=u$. Interchanging the two congruences similarly expresses $v$ as a power of $g$. Consequently $\mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }v\mathrm{\mid }$ divide $\mathrm{\mid }g\mathrm{\mid }$, so $\mathrm{\mid }u\mathrm{\mid }\mid p^a$ and $\mathrm{\mid }v\mathrm{\mid }\mid m$.

Using the exponent from step 1.1 in the factorization $g=uv$ gives $g^{sm}=u^{sm}{v}^{sm}=u,$ because $v^m=1$ and $sm\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{p}^a)$, hence modulo $\mathrm{\mid }u\mathrm{\mid }$. Likewise $g^{r{p}^a}=v$. Thus $u=g_p$ and $v=g_{p^{\mathrm{\prime }}}$, proving uniqueness.

For $h\in G$, the pair $(h{g}_p{h}^{-1},h{g}_{p^{\mathrm{\prime }}}{h}^{-1})$ is a commuting $p$-by-$p^{\mathrm{\prime }}$ factorization of $hg{h}^{-1}$. Uniqueness therefore identifies it with $((hg{h}^{-1}{)}_p,(hg{h}^{-1}{)}_{p^{\mathrm{\prime }}})$. If $a=0$, the formulas give $g_p=1,g_{p^{\mathrm{\prime }}}=g$; if $m=1$, they give $g_p=g,g_{p^{\mathrm{\prime }}}=1$. Thus the endpoint cases are included.

By F2, restriction to $H$ gives the finite decomposition ${\mathrm{Res}}_H^G\chi ={\sum }_{\zeta }{n}_{\chi ,\zeta }\zeta$. Since $u\in Z(H)$, its representing operator commutes with $H$; F3 and the splitting condition give ${\rho }_{\zeta }(u)={\lambda }_{u,\zeta }\mathrm{id}$. Moreover ${\rho }_{\zeta }(u{)}^{\mathrm{\mid }u\mathrm{\mid }}=1$, so ${\lambda }_{u,\zeta }^{\mathrm{\mid }u\mathrm{\mid }}=1$ without any algebraic-closedness assumption.

Let $v\in H$ be $p$-regular. Because $u$ and $v$ commute, trace gives $\zeta (uv)=\mathrm{tr}({\rho }_{\zeta }(u){\rho }_{\zeta }(v))={\lambda }_{u,\zeta }\zeta (v)\mathrm{.}$ By F4, $\zeta (v)={\sum }_{\phi }{d}_{\zeta ,\phi }\phi (v)$. Substitution into the restriction formula and reordering its finite sums yields $\chi (uv)={\sum }_{\phi }\left({\sum }_{\zeta }{n}_{\chi ,\zeta }{\lambda }_{u,\zeta }{d}_{\zeta ,\phi }\right)\phi (v)\mathrm{.}$ The coefficient in parentheses is exactly $d_{\chi ,\phi }^u$ by F1.

Let $v_1,\dots ,v_r$ represent the $p$-regular conjugacy classes of $H$, and list the irreducible Brauer characters as ${\phi }_1,\dots ,{\phi }_r$; equality of the two cardinalities follows from [F6]. By [F5], the evaluation matrix $M=({\phi }_j(v_i){)}_{i,j}\in M_r(K)$ has entries in the cyclotomic subfield generated by the Teichmüller lifts that occur. Applying the standard complex realization entrywise gives the Brauer-character table from [F6], whose determinant is nonzero by complex linear independence. Hence $\det M\mathrm{\ne }0$ already in that cyclotomic subfield, and therefore $M$ is invertible over $K$. If another family in $K^r$ gave the same values, subtraction would yield $Ma=0$; invertibility forces $a=0$. This proves uniqueness without applying a complex-linear independence assertion directly to $K$-coefficients.

If $u=1$, then $H=G$, the restriction has only the constituent $\chi$ with multiplicity one, and ${\lambda }_{1,\chi }=1$. The formula in F1 becomes $d_{\chi ,\phi }^1=d_{\chi ,\phi }$. The case $v=1$ in step 2.1 is valid and gives the corresponding value at $u$. All decompositions and sums are finite, so no choice principle is used.

Put $A=\mathcal{O}H$. It is finite free over $\mathcal{O}$, hence complete and separated for the $\mathfrak{m}$-adic topology by F1. Every element of $\mathfrak{m}A$ lies in $J(A)$: if $x\in \mathfrak{m}A$ and $y\in A$, then $(yx{)}^n\to 0$, so $1-yx$ has inverse ${\sum }_{n\ge 0}(yx{)}^n$. This is the Jacobson-radical test.

Let $\bar{e}\in A\mathrm{/}\mathfrak{m}A=kH$ be idempotent and choose any lift $e_1\in A$. Inductively, if $e_n^2-e_n\in {\mathfrak{m}}^{n}A$, apply the Newton correction $e_{n+1}=e_n-(2e_n-1)(e_n^2-e_n)\mathrm{.}$ Because $e_n$ commutes with $e_n^2-e_n$, direct expansion shows that $e_{n+1}^2-e_{n+1}$ is a multiple of $(e_n^2-e_n{)}^2$, while $e_{n+1}-e_n\in {\mathfrak{m}}^{n}A$. Thus the errors tend to zero and $(e_n)$ is Cauchy. Completeness gives a limit $e$ with $e^2=e$ and reduction $\bar{e}$.

Suppose now that $\bar{e}$ is central. For $X=eA(1-e)$ its reduction is $\bar{e}(kH)(1-\bar{e})=0$, hence $X=\mathfrak{m}X$: the nontrivial inclusion uses that $\mathfrak{m}=(\pi )$ in the DVR, since $x=\pi a=e(\pi a)(1-e)=\pi (ea(1-e))$. The finite $\mathcal{O}$-module $X$ has generators $x_1,\dots ,x_t$ with $x_i=\pi {\sum }_j{a}_{ij}{x}_j$. Multiplying $(I-\pi (a_{ij}))(x_j)=0$ by its adjugate shows that $\det (I-\pi (a_{ij}))x_j=0$ for every $j$. The determinant is congruent to $1$ modulo $\mathfrak{m}$, hence is a unit, so $X=0$. Applying the same argument to $(1-e)Ae$ gives that space zero too. Therefore $ea=eae=ae$ for every $a\in A$, and $e$ is central.

If central idempotents $e,f\in A$ have the same reduction, then $e-f\in \mathfrak{m}A\subseteq J(A)$. The commuting products $e(1-f)$ and $f(1-e)$ are idempotents in $J(A)$, and an idempotent in the Jacobson radical is zero. Hence $e=ef=f$. Thus every central idempotent of $kH$ has exactly one central lift.

Central primitivity is preserved. If a central lift $e$ decomposed into two nonzero orthogonal central idempotents, neither summand could reduce to zero, since step 1.1 puts its kernel inside $J(A)$; their reductions would decompose $\bar{e}$. Conversely, a central decomposition of $\bar{e}$ lifts termwise by steps 2.1–3.1, and uniqueness in step 4.1 makes the lifted sum equal to $e$. Hence $e$ is primitive exactly when $\bar{e}$ is primitive. Together with F2 this proves the bijection and the asserted unique block lift, including the trivial-group case. The constructions are finite or sequential limits fixed by explicit formulas, so no choice principle is used.

Let $M$ be such a lattice and put $E={\mathrm{End}}_{\mathcal{O}H}(M)$. As an $\mathcal{O}$-submodule of the finite free module ${\mathrm{End}}_{\mathcal{O}}(M)$, the module $E$ is finite free and $\mathfrak{m}$-adically complete. Also $\mathfrak{m}E\subseteq J(E)$: for $x\in \mathfrak{m}E$ and $y\in E$, the geometric series ${\sum }_{n\ge 0}(yx{)}^n$ converges and inverts $1-yx$.

The algebra $\bar{E}=E\mathrm{/}\mathfrak{m}E$ is finite-dimensional over $k$, so its Jacobson radical $J(\bar{E})$ is nilpotent and $\bar{E}\mathrm{/}J(\bar{E})$ is semisimple Artinian. Because $\mathfrak{m}E\subseteq J(E)$ by step 1.1, the standard quotient identity gives $J(E\mathrm{/}\mathfrak{m}E)=J(E)\mathrm{/}\mathfrak{m}E\mathrm{.}$ For completeness, if $x+\mathfrak{m}E$ lies in the radical of the quotient, then for every $y\in E$ the element $1-yx$ is a unit modulo $\mathfrak{m}E$; lifting a two-sided inverse leaves errors in $\mathfrak{m}E\subseteq J(E)$, and multiplying by the inverses of $1$ minus those errors gives a two-sided inverse in $E$. Thus $x\in J(E)$ by the Jacobson-radical test. The reverse inclusion follows by passing units to the quotient. Consequently $E\mathrm{/}J(E)\cong \bar{E}\mathrm{/}J(\bar{E})$ is semisimple Artinian.

Idempotents lift from $E\mathrm{/}J(E)$ to $E$. First lift through the nilpotent ideal $J(\bar{E})$: successively through its powers, the polynomial Newton correction to $e^2-e$ turns an error in $J(\bar{E}{)}^n$ into one in $J(\bar{E}{)}^{2n}$, so the finite nilpotence filtration terminates. Then lift the resulting idempotent of $E\mathrm{/}\mathfrak{m}E$ by the same corrections in $E$; completeness makes the corrections converge. Therefore $E$ is semiperfect.

A nontrivial idempotent of $E$ is exactly a nontrivial direct-sum decomposition of $M$. Repeatedly splitting such an idempotent terminates, because the positive $\mathcal{O}$-ranks of both summands are smaller; direct summands remain finite free over the DVR. This proves existence of a finite indecomposable decomposition. If $M$ is indecomposable, $E$ has no nontrivial idempotent. By step 3.1, every idempotent of the semisimple Artinian ring $E\mathrm{/}J(E)$ lifts, so that quotient also has no nontrivial idempotent. It must therefore be a division ring. Hence the nonunits of $E$ are exactly $J(E)$, and $E$ is local.

Suppose $M={⨁}_{i=1}^r{M}_i={⨁}_{j=1}^s{N}_j$ are two indecomposable decompositions. Restrict the identity of $M$ to $M_1$ through the second decomposition. It becomes a finite sum of composites $M_1\to N_j\to M_1$. In the local ring ${\mathrm{End}}_{\mathcal{O}H}(M_1)$, a sum of nonunits cannot be $1$; therefore one composite is a unit. Write that composite as $ab$, where $b:M_1\to N_j$ is the second-decomposition projection restricted to $M_1$ and $a:N_j\to M_1$ is the first-decomposition projection restricted to $N_j$. Replacing $b$ by $b(ab{)}^{-1}$ gives $ab=1_{M_1}$. Hence $N_j=b(M_1)\oplus \ker a$; indecomposability and $b(M_1)\mathrm{\ne }0$ force $\ker a=0$, so $a$ is an isomorphism. Relative to $M=M_1\oplus M^{\mathrm{\prime }}$ with $M^{\mathrm{\prime }}={⨁}_{i>1}{M}_i$, the summand $N_j$ is therefore the graph of a map $M_1\to M^{\mathrm{\prime }}$. Subtracting that graph map is an automorphism of $M$ which fixes $M^{\mathrm{\prime }}$ and carries $N_j$ to $M_1$. Thus $M=N_j\oplus M^{\mathrm{\prime }}$, and quotienting by $N_j$ identifies $M^{\mathrm{\prime }}$ with the sum of the remaining $N$-summands. Induction on the rank matches all summands and their multiplicities. The zero lattice has the empty decomposition, while the local ring assertion was stated only for nonzero indecomposables. Every selection is from a finite decomposition, so no choice principle is used.

Decompose $H$ into its finite $A$-$B$ double cosets. The summand of $\mathcal{O}H{\otimes }_{\mathcal{O}B}V$ supported on $AxB$ is identified with $\mathcal{O}A{\otimes }_{\mathcal{O}{L}_x}{}^{x}V,a\otimes v⟼ax\otimes v,$ where $q\in L_x$ acts on ${}^{x}V$ as $x^{-1}qx$ acts on $V$. These maps and their inverses are well-defined on the tensor relations, and their finite direct sum is the displayed Mackey isomorphism. Tensor associativity gives ${\mathrm{Ind}}_A^H{\mathrm{Ind}}_B^A\cong {\mathrm{Ind}}_B^H$ for $B\le A\le H$. Since $\mathcal{O}H$ is finite free as a right subgroup algebra, these operations preserve finite-free lattices; functoriality preserves split inclusions and retractions.

First connect F1's summand definition to a split induction counit. Put $Y={\mathrm{Ind}}_Q^{H}V=\mathcal{O}H{\otimes }_{\mathcal{O}Q}V$. The counit ${\epsilon }_Y:{\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}Y\to Y$ has the explicit $H$-linear section $s_Y(h\otimes v)=h\otimes (1\otimes v)\mathrm{.}$ It respects the relation $hq\otimes v=h\otimes qv$ because $q\otimes v=1\otimes qv$ in $Y$, and ${\epsilon }_Y{s}_Y=1_Y$. If $M$ is a summand of $Y$ with inclusion $i:M\to Y$ and retraction $r:Y\to M$, then $s_M=({\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}r)s_{Y}i$ splits the counit for $M$, by naturality of the counit and $ri=1_M$. Conversely, a split counit displays $M$ as a summand of its induced module.

The image of every relative trace is a two-sided ideal of $E={\mathrm{End}}_{\mathcal{O}H}(M)$: an $H$-endomorphism can be moved inside either side of the finite trace sum. Suppose now that $M$ is indecomposable, relatively $P$-projective and relatively $R$-projective. By step 1.2 choose trace expressions for $1_M$ from $P$ and from $R$ and multiply them. The diagonal $H$-orbits on the finite set $H\mathrm{/}P\times H\mathrm{/}R$ regroup the product as a finite sum of relative traces from their stabilizers $sP{s}^{-1}\cap tR{t}^{-1}\mathrm{.}$ The element inside each orbit trace is fixed by that stabilizer, so every summand belongs to the corresponding trace ideal.

By F2 the ring $E$ is local. If every summand from step 2.1 were a nonunit, their sum could not be $1_M$; hence one is a unit. Its two-sided trace ideal then contains $1_M$, and step 1.2 makes $M$ relatively projective for the associated intersection. Conjugating that subgroup by $s^{-1}$ shows that $M$ is relatively $P\cap s^{-1}tR{t}^{-1}s$, a subgroup of $P$. If $P$ is a vertex, its minimality forces this intersection to be $P$. Thus $P\le s^{-1}tR{t}^{-1}s$, as required. Empty double-coset sets cannot occur because $H$ is nonempty; trivial subgroups and $P=1$ are included. All coset sets and sums are finite, so no choice principle is used.

Put $X={\mathrm{Ind}}_N^{H}L$ and let $I_H(L)=\{h\in H:{}^{h}L\cong L\}$. This is a subgroup containing $N$, so the prime-index hypothesis gives $I_H(L)=N$ or $H$. By F1, $E_0={\mathrm{End}}_{\mathcal{O}N}(L)$ is local. Its residue division ring $E_0\mathrm{/}J(E_0)$ is finite-dimensional over $k$; F3 makes it equal to $k$, because every element satisfies a split polynomial and a division ring has no nonzero zero divisors.

Suppose $I_H(L)=N$. On restriction to $N$, $X$ is the direct sum of the $p$ pairwise nonisomorphic conjugates of $L$. Krull--Schmidt and step 1.1 identify the semisimple quotient of its $N$-endomorphism ring with $k^p$: diagonal entries reduce modulo the local radicals, while every map between distinct indecomposable summands belongs to the categorical radical. Conjugation by a generator of $H\mathrm{/}N$ cyclically permutes these $p$ factors. An $H$-endomorphism idempotent therefore has image $(0,\dots ,0)$ or $(1,\dots ,1)$ in $k^p$. In the first case the idempotent lies in the Jacobson radical and is zero; in the second its complement does, so it is one. Thus $X$ has no nontrivial $H$-endomorphism idempotent and is indecomposable.

Suppose $I_H(L)=H$. Choose isomorphisms among the $p$ conjugate summands. They identify the semisimple quotient of ${\mathrm{End}}_{\mathcal{O}N}(X)$ with $M_p(k)$. The action of a generator of $H\mathrm{/}N$ on this quotient is conjugation by a matrix $T$ that cyclically permutes the $p$ diagonal primitive idempotents. Its $p$th power is scalar, say $T^p=\lambda I$. By F3 choose $\mu \in k$ with ${\mu }^p=\lambda$; in characteristic $p$, $z^p-\lambda =(z-\mu {)}^p$. The cyclic permutation of the diagonal idempotents makes $T$ cyclic of degree $p$, so its Jordan form is one block and $C_{M_p(k)}(T)=k[T]\cong k[z]\mathrm{/}((z-\mu {)}^p),$ a local algebra.

Every $H$-endomorphism of $X$ is an $N$-endomorphism fixed by this conjugation. Hence an $H$-endomorphism idempotent maps to an idempotent in the local centralizer computed in step 2.2, and that image is $0$ or $1$. The kernel of the reduction to $M_p(k)$ lies in the Jacobson radical of the $N$-endomorphism ring; an idempotent in it is zero, and the same argument applied to the complement handles image $1$. Thus again only $0$ and $1$ occur, so $X$ is indecomposable. The two inertia cases are exhaustive. The nonzero hypothesis excludes the zero lattice, and the proof uses only finite decompositions; algebraic closedness is used exactly in steps 1.1 and 2.2, not as a hidden choice principle.

Since $Z$ is central, $C_H(Z)=H$. Consequently F1 gives ${\mathrm{Br}}_Z(c)=c\mathrm{.}$ The primitive idempotent $c$ is nonzero, so F2 says that for every defect group $D$ of $c$ there is an $h\in H$ such that $Z\le hD{h}^{-1}$.

Conjugating this containment by $h^{-1}$ gives $h^{-1}Zh\le D$. Centrality gives $h^{-1}Zh=Z$, and hence $Z\le D$. This includes $Z=1$ and applies to each defect group separately.

The assertion is immediate if $M=0$, so assume otherwise. Let $t$ be the $p$-part of $x$. Since $1\in Q$, the hypothesis implies $t\mathrm{\ne }1$; in particular $p$ divides $\mathrm{\mid }x\mathrm{\mid }$. Put $C=⟨x⟩\text{and}L=⟨x^p⟩\mathrm{.}$ Then $L⊴C$ and $[C:L]=p$. The subgroup generated by $t$ is the Sylow $p$-subgroup of the cyclic group $C$. If $R\le C$ does not contain $t$, then $R\le L$: indeed, any power $x^j\mathrm{\notin }L$ has $p\nmid j$ and generates the whole Sylow $p$-subgroup of $C$.

By F1, $M$ is a direct summand of $Y={\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}M$. Restricting to $C$ and using F2 gives ${\mathrm{Res}}_C^{H}Y\cong {⨁}_{s\in C\mathrm{\backslash }H\mathrm{/}Q}{\mathrm{Ind}}_{R_s}^C{V}_s,R_s=C\cap sQ{s}^{-1},$ for the corresponding integral $R_s$-lattices $V_s$. If some $R_s$ contained $t$, then $t$ would belong to the $H$-conjugate $sQ{s}^{-1}$, contrary to the hypothesis. Thus step 1.1 gives $R_s\le L$ for every $s$. By induction transitivity, ${\mathrm{Ind}}_{R_s}^C{V}_s\cong {\mathrm{Ind}}_L^C\left({\mathrm{Ind}}_{R_s}^L{V}_s\right)\mathrm{.}$

Decompose each nonzero $\mathcal{O}L$-lattice ${\mathrm{Ind}}_{R_s}^L{V}_s$ into indecomposables $W_{s,j}$ by F3. Because $L⊴C$ has index $p$, F4 says that every ${\mathrm{Ind}}_L^C{W}_{s,j}$ is indecomposable. Hence step 2.1 is an indecomposable decomposition of ${\mathrm{Res}}_C^{H}Y$. Since ${\mathrm{Res}}_C^{H}M$ is a direct summand, Krull--Schmidt makes each of its indecomposable summands isomorphic to one of these induced lattices.

For any $\mathcal{O}L$-lattice $W$, scalar extension identifies $K{\otimes }_{\mathcal{O}}{\mathrm{Ind}}_L^{C}W$ with ${\mathrm{Ind}}_L^C(K{\otimes }_{\mathcal{O}}W)$. Its direct-sum decomposition over the $p$ cosets of $L$ in $C$ is cyclically permuted by $x$, because $x\mathrm{\notin }L$. Thus the matrix of $x$ has zero diagonal blocks, and its trace is zero. Applying this to every summand selected in step 3.1 and adding their traces yields $\theta (x)=\mathrm{tr}(x\mid K{\otimes }_{\mathcal{O}}M)=0.$ Algebraic closedness is used exactly through Green indecomposability in step 3.1; all selections and sums are finite.

Since $H\le N_G(D)$, the group $D$ is normal in $H$. Hence $D\le O_p(H)$, and F6 shows that every defect group $Q$ of every block $c$ of $kH$ contains $D$. It follows that $C_G(Q)\le C_G(D)\le H\mathrm{.}$ Thus F5 defines $c^G$ for every block $c$ of $kH$.

We record the block corner that controls an error component. Write $b$ for the idempotent of the global block $B$, and let ${\pi }_H:kG\to kH$ delete coefficients outside $H$. For a block $c$ of $kH$, the maps $kHc\hookrightarrow kG,a⟼c{\pi }_H(a):kG⟶kHc$ split the identity $H$-double-coset copy of the block bimodule $kHc$. The corner of left multiplication by $b$ on this copy is left multiplication by $c{\pi }_H(b)$. If that corner were a unit in ${\mathrm{End}}_{H\times H}(kHc)=Z(kHc)$, normalizing the second map by its inverse would split $kHc$ from the restriction of the global block $kGb$. By F5 this would imply $c^G=B$. Therefore, when $c^G\mathrm{\ne }B$, the element $c{\pi }_H(b)$ is a nonunit of the local algebra $Z(kHc)$ and hence is nilpotent.