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Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

Throughout, modules are finite-dimensional over $k$, or finitely generated $\mathcal{O}$-free lattices when the coefficient ring is $\mathcal{O}$. The p-modular system has a complete discrete valuation ring $\mathcal{O}$, maximal ideal $(\pi )$, fraction field $K$ of characteristic zero and residue field $k$ of characteristic $p$.

A block of a group algebra is a nonzero, indecomposable two-sided ideal direct summand of $A=RG$. Equivalently it is $Ae$ for a primitive central idempotent $e$. To see the equivalence, suppose $A=B\oplus C$ as two-sided ideals and write $1=e+f$ with $e\in B$, $f\in C$. Products between $B$ and $C$ vanish because they lie in their zero intersection. Thus $e$ is the identity on $B$, zero on $C$, and is a central idempotent, with $B=Ae$. Conversely $A=Ae\oplus A(1-e)$. Splitting $Ae$ further is equivalent to splitting $e$ into orthogonal nonzero central idempotents; indecomposability is exactly primitivity. If $1=\sum e_i$ is the block decomposition, $M=⨁e_{i}M$. A nonzero indecomposable module therefore belongs to exactly one block of a group algebra, namely the one with $e_{i}M=M$.

For modular reduction of block idempotents, the conjugacy class sums are an $R$-basis of $Z(RG)$. Hence $Z(\mathcal{O}G)/\pi Z(\mathcal{O}G)\cong Z(kG)$. In the complete commutative algebra $Z(\mathcal{O}G)$, idempotent lifting gives a unique lift of each idempotent of $Z(kG)$; uniqueness follows because commuting idempotents congruent modulo the Jacobson radical are equal. Orthogonal decompositions lift too, so primitivity is preserved. Reduction gives a bijection between the integral and modular blocks.

Here is a definition of defect groups of a block which also proves the required relative-projectivity assertion. Give $RG$ its conjugation action and, for $H\le G$, define the relative trace and its transfer ideal of conjugation-fixed elements byA defect group of a block $RGe$ is a $p$-subgroup $D$ minimal with $e\in I_D$. Such a subgroup exists: for a Sylow subgroup $P$, $[G:P]$ is a unit in $R$ and $e={\mathrm{Tr}}_P^G(e/[G:P])$. The defect of a block is $d$ where $|D|=p^d$.

For completeness, this definition is independent of a choice up to conjugacy. Multiplying two relative traces and grouping the pairs of cosets by double cosets givesThis identity follows by writing each pair of cosets as a simultaneous $G$-orbit; its stabilizer is $D\cap {}^{g}E$, and its contribution is a trace from that stabilizer. If $D,E$ are both minimal for $e$, then $e=e^2$ belongs to the displayed sum. The commutative block center $eZ(RG)$ is a local ring: in the field case it is a commutative Artinian ring with no nontrivial idempotents; in the integral case its reduction has this property and completeness detects units. A sum of proper ideals cannot contain its identity. Consequently $e\in I_{D\cap {}^{g}E}$ for some $g$, forcing $D={}^{g}E$ by minimality. The same argument with an arbitrary $E$ shows that $e\in I_E$ precisely when $E$ contains a conjugate of $D$. This is the trace criterion for defect groups.

To relate this to the usual modular definition, the Brauer morphism deletes the coefficients outside the centralizer:It is an algebra homomorphism: in the coefficient of an element centralizing $P$ in a product, $P$ acts on the contributing pairs; nonfixed orbits have size divisible by $p$, while fixed pairs have both entries in $C_G(P)$. Its kernel is the sum of proper-subgroup transfers, because the nonfixed conjugation orbit sums form a basis of that kernel. If $D$ is minimal for $e={\mathrm{Tr}}_D^G(a)$ and ${\mathrm{Br}}_D(e)=0$, replace $a$ by $ea$. The kernel description gives $e\in {\sum }_{Q<D}{I}_Q$, contradicting the local-center argument. Also ${\mathrm{Br}}_P(I_D)=0$ unless $P$ is conjugate into $D$: group the cosets of $G/D$ into $P$-orbits, which have no fixed point otherwise. Thus the minimal transfer subgroups are exactly the maximal $p$-subgroups with nonzero Brauer morphism of $e$.

Now take $M=eM$ and write $e={\mathrm{Tr}}_D^G(a)$. Multiplication by $a$ is a $D$-linear endomorphism $\alpha$ of $M$, andHere is the splitting behind the D. Higman criterion. The natural map $\epsilon :RG{\otimes }_{RD}M\to M$, $g\otimes m\mapsto gm$, has a $G$-linear sectionChanging a representative by an element of $D$ leaves this tensor unchanged, $G$ permutes the terms, and $\epsilon s={\mathrm{id}}_M$. HenceThis is relative projectivity for $D$, proving the Green relative projectivity theorem; indecomposability was needed only to speak of a single block of a group algebra or vertex of an indecomposable module.

Finally let $\hat{e}$ lift $e$. A trace expression for $\hat{e}$ reduces to one for $e$, so a modular defect group of a block is contained, up to conjugacy, in an integral one. Conversely choose a modular defect group $D$ and $a\in (kG{)}^D$ with $e={\mathrm{Tr}}_D^G(a)$. Conjugation orbit sums lift $a$ to $\hat{a}\in (\mathcal{O}G{)}^D$. The elementreduces to its block identity, hence is a unit in that block center. Its inverse there is central, soThe integral defect is therefore contained in the modular one. The corresponding blocks have the same conjugacy class of defect groups and the same defect.

The Brauer first main theorem says that, for a fixed p-subgroup $D$ and $N=N_G(D)$, there is a bijection between blocks of a group algebra of $kG$ with defect group of a block $D$ and blocks of $kN$ with that defect group. Correspondents have the same nonzero image under ${\mathrm{Br}}_D$. The integral version follows by modular reduction of block idempotents.

Here is a proof using relative traces. Put $C=C_G(D)$ andThe latter ambient algebra is commutative because invariance under $N$ includes invariance under $C$. The Brauer morphism and relative trace identity isIndeed $D$ acts on $G/D$. Its fixed cosets are exactly $N/D$. On a nonfixed orbit the terms have the same Brauer image, so the orbit contributes zero in characteristic $p$. The fixed cosets give the right-hand side. Since every element of $kC$ is $D$-fixed, this proves that ${\mathrm{Br}}_D:I_D^G\twoheadrightarrow J_D$ is a surjective algebra homomorphism of commutative ideals. The same argument for $N$ has exactly the same target.

We use the following elementary fact about primitive idempotents under a surjection of commutative Artinian ideals. Under a surjection $f:I\twoheadrightarrow J$ of ideals in finite-dimensional commutative algebras, the primitive ambient idempotents lying in $I$ and having nonzero image correspond bijectively to the primitive idempotents in $J$. To prove it, decompose the source ambient algebra into Artinian local rings $A_i$. Either $I\cap A_i=A_i$, or it is a proper, hence nilpotent, ideal. Proper components contribute no idempotents to the image. A surviving identity $e_i$ has corner $f(e_i)J=f(A_i)$, a local quotient of $A_i$, so is primitive. The surviving identities are orthogonal and exhaust the idempotents of the target. Notice that this proof does not assert that the whole kernel is nilpotent: it can kill entire block components.

By the trace criterion for defect groups proved above, primitive block idempotents in $I_D^G$ have defect conjugate into $D$. Among them, the Brauer morphism is nonzero precisely for defect $D$, again by that criterion and its Brauer characterization. The identical reasoning applies to $I_D^N$. Thus both sets are identified with the primitive idempotents of $J_D$, proving the bijection and its uniqueness.

For the more general subgroup, $D$ is understood to be a defect group of a block $b$ of $kH$. This is the implicit hypothesis needed to apply the theorem; it is not repeated in the printed sentence. Without it the printed uniqueness assertion is false: take $G=A_5$, $H=A_4$, $p=3$ and $D=C_3$. The centralizer of $D$ in $A_5$ is $D$, but the unique three-dimensional defect-zero block of $k{A}_4$ is the bimodule restriction of each of the two three-dimensional defect-zero blocks of $k{A}_5$. The two ordinary degree-three characters both restrict to the degree-three character of $A_4$, with values $3,-1,0$ on its identity, involutions and elements of order three. Defect-zero reduction gives the same simple projective $k{A}_4$-module, so both restricted block bimodules are this same matrix block. This illustrates why block induction requires the block's defect centralizer. Write $e_b$ for its block idempotent and $K=N_H(D)$. Since $D{C}_G(D)\le H$, we have $D{C}_G(D)\le K\le N_G(D)$. Apply the theorem inside $H$ to obtain the unique corresponding block $\beta$ of $kK$. Its center of an associative algebra has a residue field $F_{\beta }$, and the quotient map is its central character of a block ${\omega }_{\beta }:Z(kK)\to F_{\beta }$. No assumption that $k$ is algebraically closed is needed here. DefineThe Brauer image is in $(k{C}_G(D){)}^{N_G(D)}\subseteq Z(kK)$, so this is a unital algebra homomorphism. Among the orthogonal primitive central idempotents of $kG$, exactly one has image $1$. The block it selects is $b^G$. If $H$ contains $N_G(D)$ this is the inverse of the bijection just proved. For arbitrary $H$, the resulting block can have a strictly larger defect group. Block induction is transitive on admissible intermediate subgroups. For the normalizer chain used in the proof below, the required transitivity will be verified directly with Brauer images.

To prove the bimodule characterization, we explicitly state the standard permutation-module fact used here. For a permutation module, the Brauer quotient of a module at $P$ has as basis the permutation-basis points fixed by $P$. An indecomposable summand has nonzero quotient at $P$ exactly when its vertex of an indecomposable module contains a conjugate of $P$. For summands with vertex exactly $P$, taking this quotient, with its $N(P)/P$-action, detects the summand and detects which orthogonal idempotent projection acts as its identity. These facts hold for trivial source modules; the fixed-basis description follows by deleting proper transfers, and the detection assertion is the permutation form of the first-main-theorem argument, applied to their endomorphism algebra.

As an $H\times H$ permutation module, with $(h_1,h_2)g=h_{1}g{h}_2^{-1}$, the group algebra decomposes asThe block bimodule $b$ is an indecomposable module with vertex of an indecomposable module $\Delta D=\{(d,d):d\in D\}$. None of the other double-coset modules can contain $b$: a basis element fixed by $\Delta D$ centralizes $D$, hence lies in $C_G(D)\subseteq H$, whereas those double cosets lie outside $H$. Thus $b$ occurs exactly once as an indecomposable summand of this restriction, in the $kH$ term. Decomposing $kG$ into its global block bimodules therefore puts this occurrence in precisely one of them, say $B$.

Under the fixed-basis description, the quotient of $kG$ at $\Delta D$ is $k{C}_G(D)$, and multiplication by $e_B$ induces multiplication by ${\mathrm{Br}}_D(e_B)$. The quotient of $b$ is ${\mathrm{Br}}_D^H(e_b)k{C}_G(D)=e_{\beta }k{C}_G(D)$. Consequently the detecting projection for this occurrence of $b$ satisfieswhich is equivalent to ${\omega }_b^G(e_B)=1$. Thus $B=b^G$, proving the requested unique-block direct-summand characterization. Here the restriction is to $H\times H$; the extra subscript in the converted TeX is a transcription error, not a different subgroup.