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If $N_G(D)\le H\le G$, Brauer correspondence pairs the blocks of a group algebra of $RG$ and $RH$ having defect group of a block $D$. This is Green correspondence for their bimodules, with vertex of an indecomposable module $\Delta D$ in $G\times G$. The normalizer of $\Delta D$ lies in $H\times H$, and the corresponding restriction and induction have unique summands with that vertex.

The block statement is Green correspondence for blocks: if $N_G(D)\le H\le G$, blocks of $RG$ and $RH$ with defect group $D$ correspond, and their block bimodules are Green correspondents for the diagonal vertex $\Delta D$. We prove the module correspondence underlying this statement, then specialize it to the block bimodules. This also gives the requested direct proof in the trivial-intersection situation.

Here are the elementary ingredients, including their proofs. Mackey decomposition is obtained by partitioning the tensor basis of $RG{\otimes }_{RL}U$ by the double cosets $H\backslash G/L$; the term for $g$ is induction from $H\cap {}^{g}L$ of the suitably conjugated restriction of $U$. The D. Higman criterion follows from the section constructed in Question 1. Conversely, take a splitting of the induction counit and its coefficient at the identity coset: equivariance reconstructs the section from that $H$-linear coefficient, and the splitting identity is exactly that its relative trace is $1$. Thus the trace criterion is an equivalence.

An indecomposable module here has a local endomorphism ring. Over $k$ this follows from the Fitting lemma: an endomorphism eventually splits its kernel from its image, and indecomposability makes it invertible or nilpotent. Over $\mathcal{O}$, the endomorphism ring is finite over a complete discrete valuation ring; lift orthogonal idempotents from its finite-dimensional residue algebra to see that an indecomposable lattice has a local endomorphism ring too. The Krull–Schmidt theorem follows by the usual exchange argument: composing the inclusion and projection of one indecomposable summand across a second decomposition writes its identity as a sum of endomorphisms; one is a unit in the local endomorphism ring and pairs that summand with a summand of the second decomposition. Remove the pair and continue. Existence follows by successively splitting nontrivial idempotents, which decreases dimension or rank.

A vertex of an indecomposable module is a subgroup minimal for relative projectivity. It can be taken to be a $p$-subgroup because averaging over the index of a Sylow subgroup of any candidate subgroup is invertible. If an indecomposable module is relatively projective for $P$ and $Q$, the Mackey decomposition and its induction splittings make it a summand of a sum of modules induced from $P\cap {}^{g}Q$. In its local endomorphism ring one splitting component must be invertible, so it is relatively projective for one of these intersections. This proves conjugacy of minimal vertices and the useful criterion that a module with vertex $P$ can be relatively projective for $Q$ only when $P$ is conjugate into $Q$. A source of an indecomposable module with vertex $P$ is an indecomposable summand $S$ of its restriction to $P$ from which the original module is an induced summand. Such a summand exists by the induction splitting and Krull–Schmidt theorem; its vertex is $P$, since a proper vertex would make the original module relatively projective for a proper subgroup of $P$.

SetWhen saying a module is $\mathcal{X}$- or $\mathcal{Y}$-projective, allow conjugates in the relevant ambient group and finite sums of modules relatively projective for members of that family. No conjugate of $D$ inside $H$ belongs to $\mathcal{Y}$: if $D\le {}^h(H\cap {}^{g}D)$, equality of orders gives $D={}^{hg}D$, so $hg\in N_G(D)\subseteq H$, contradicting $g\notin H$. Every member of $\mathcal{X}$ is a proper subgroup of $D$.

The two key assertions can now be proved rather than assumed. First, if $W$ is an indecomposable $D$-projective $RH$-module, thenChoose $U$ with $W\mid {\mathrm{Ind}}_D^{H}U$. Mackey decomposition gives ${\mathrm{Res}}_H^G{\mathrm{Ind}}_D^{G}U={\mathrm{Ind}}_D^{H}U\oplus T$, with $T$ $\mathcal{Y}$-projective. Write ${\mathrm{Ind}}_D^{H}U=W\oplus W^{\prime \prime }$ and apply induction and restriction. Both $\mathrm{Res}\mathrm{Ind}W$ and $\mathrm{Res}\mathrm{Ind}{W}^{\prime \prime }$ contain their original modules, by the identity double coset. Comparing multiplicities of every indecomposable which is not $\mathcal{Y}$-projective, their extra multiplicities are nonnegative and have sum zero. Therefore neither has such an extra summand, proving the assertion.

Second, suppose $W$ has vertex $D$, and choose an indecomposable $V\mid {\mathrm{Ind}}_H^{G}W$ whose restriction contains $W$; such a $V$ exists by the first assertion. Write $\mathrm{Ind}W=V\oplus V^{\prime }$. Its restriction has just one summand isomorphic to $W$ and all other summands are $\mathcal{Y}$-projective, so $\mathrm{Res}{V}^{\prime }$ is $\mathcal{Y}$-projective. Any indecomposable $Z\mid V^{\prime }$ has a vertex $P\le D$ after conjugacy. A source $S$ with vertex $P$ is a summand of ${\mathrm{Res}}_{P}Z$. Restricting a $\mathcal{Y}$-projective decomposition of ${\mathrm{Res}}_{H}Z$ to $P$ and using Mackey decomposition forces $P$ to lie in an $H$-conjugate of some $H\cap {}^{g}D$, $g\notin H$. Since $P\le D$, this puts $P$ in $\mathcal{X}$. Thus $V^{\prime }$ is $\mathcal{X}$-projective. Moreover $V$ has vertex $D$, since it is $D$-projective and its restriction contains $W$ with vertex $D$. We have provedIn particular $V$ is unique.

Conversely, let $V$ be an indecomposable $RG$-module with vertex $D$ and source $S$. Decompose ${\mathrm{Ind}}_D^{H}S$ and choose $W$ whose induction contains $V$. Its vertex is contained in $D$ and cannot be proper, since induction would then make $V$ relatively projective for a smaller subgroup. Thus $W$ has vertex $D$, and the previous assertion makes $V$ its unique vertex-$D$ induced summand. Restriction of an $\mathcal{X}$-projective module is $\mathcal{Y}$-projective: in a Mackey term a conjugating element inside $H$ preserves the containment in an $H$-conjugate of $\mathcal{Y}$, and one outside $H$ gives a subgroup of $H\cap {}^{g}D$. Comparing the two decompositions of $\mathrm{Res}\mathrm{Ind}W$ now givesThese assignments are mutually inverse. To check sources, restrict ${\mathrm{Ind}}_D^{H}S$ back to $D$. Its terms of full vertex are conjugates of $S$ by $N_H(D)$; the other terms are induced from proper intersections. Applying the source criterion and Krull–Schmidt theorem shows that the source of $W$ is one of those conjugates. The same argument in $G$, and the uniqueness just proved, matches the source classes on the two sides. Thus Green correspondence preserves sources up to normalizer conjugacy.

To pass to blocks of a group algebra, regard $RG$ as the permutation module $R[(G\times G)/\Delta G]$. A summand of a permutation module has trivial source. Here is why: over a finite p-group, every transitive permutation module is indecomposable, since its socle is the one-dimensional fixed line and every nonzero module has nonzero fixed points. The latter follows by induction on the group order, using a central element of order $p$ and the nilpotence of its difference from $1$. Over $\mathcal{O}$ indecomposability follows by reduction. Restrict a permutation summand to a vertex $P$ and apply Krull–Schmidt theorem; a source must be a transitive permutation module $R[P/Q]$. Its vertex is contained in $Q$, so minimality forces $Q=P$, and its source is the trivial module.

For a permutation module, its Brauer quotient of a module at $P$ has the fixed permutation points as basis: every nontrivial orbit sum is a proper transfer. An indecomposable trivial-source summand with vertex $Q$ has nonzero quotient only at subgroups conjugate into $Q$, by its induced-permutation realization, and its quotient at $Q$ is nonzero because its restriction contains its trivial source. Applied to a block $B=kGe$, the quotient at $\Delta P$ is ${\mathrm{Br}}_P(e)k{C}_G(P)$. Together with the maximal-subgroup characterization of defect from Question 1, this proves that its bimodule vertices are precisely $\Delta D$ for its defect groups $D$. The integral conclusion follows by orbit-sum reduction and idempotent lifting.

FinallyApply the just-proved Green correspondence with ambient group $G\times G$. Its unique vertex-$\Delta D$ summand of $B{\downarrow }_{H\times H}$ lies in $RH$: outside $H$ there is no $\Delta D$-fixed group-basis point, as in Question 2. Since the indecomposable $H\times H$ summands of $RH$ are exactly its block ideals, this summand is a block $b$ with defect group $D$. Conversely induction of $b$ has $B$ as its unique vertex-$\Delta D$ summand. The block bijection is the Brauer correspondence of Question 2, giving the claimed block form over either coefficient ring.

For the final special case put $N=N_G(D)$ with $D$ a trivial-intersection Sylow subgroup. For $g\notin N$, the intersection $N\cap {}^{g}N$ has order prime to $p$. Indeed every $p$-subgroup of it lies both in the normal Sylow subgroup $D$ of $N$ and in ${}^{g}D$, whose intersection is trivial. Mackey decomposition therefore giveswhere every other double-coset term is induced from a group of order prime to $p$. By Maschke's theorem such modules are projective, and induction preserves projectivity, so $P$ is projective.

Every nonprojective $kG$-module has nonprojective restriction to $N$: $[G:N]$ is prime to $p$, so averaging makes the module a summand of the induction of its restriction. A projective restriction would therefore make the original module projective. Decompose ${\mathrm{Ind}}_N^{G}M$ into indecomposables. Each nonprojective summand contributes a nonprojective summand on restriction, whereas the displayed Mackey formula and Krull–Schmidt theorem have exactly one nonprojective summand, $M$, with multiplicity one. There must be one, since $M$ is nonprojective. Hence

Write $A=kG$, $N=N_G(D)$ and $C=C_G(D)$. The Brauer morphism is the coefficient projectionwhere the superscript $D$ denotes invariance under conjugation. This is an algebra homomorphism: for $c\in C$, the pairs $(x,y)$ with $xy=c$ are permuted by $D$; their coefficient products are constant on each orbit. Every nonfixed orbit has size divisible by $p$ and contributes zero. The fixed pairs are exactly $x,y\in C$, giving the coefficient of $c$ in the product of the projections. It is surjective since $kC\subset A^D$.

We need two trace facts. The kernel isIndeed, a basis of $A^D$ consists of conjugation-orbit sums. Nonfixed orbit sums are traces from their proper stabilizers, while fixed basis elements are exactly the elements of $C$. Traces from proper subgroups have zero coefficient at every fixed element, proving the reverse inclusion. Second, Brauer morphism and relative trace givesOne can see this by letting $D$ act on the cosets $G/D$: fixed cosets are exactly $N/D$, and all other orbit contributions vanish after projection. More generally, ${\mathrm{Br}}_E^G({\mathrm{Tr}}_D^G(A^D))=0$ unless $E$ is conjugate into $D$.

Set $I_D={\mathrm{Tr}}_D^G(A^D)$ and $J_D={\mathrm{Tr}}_D^N(kC)=(kC{)}_D^N$. The first is an ideal of the center of an associative algebra $Z(A)$; the second is an ideal of $(kC{)}^N$, which is commutative because $C\subset N$. The displayed trace identity and surjectivity of the Brauer projection give a surjective algebra homomorphismThe ideals can lack identities, so we justify the required idempotent argument. A finite-dimensional commutative algebra is a product of Artinian local rings. Its ideal intersects each local factor either in the whole factor, containing its identity, or in a proper nilpotent ideal. Under a surjection of such ideals, the proper parts have nilpotent image. Each nonzero image of a local-factor identity has a local corner algebra, a quotient of that factor, and is therefore primitive. If $u$ is the sum of these nonzero images, then $j-uj$ lies in the nilpotent image of the proper-factor ideals for every $j$ in the target; an idempotent has $j-uj=0$. Its components in the local corners are consequently zero or the corner identities. These orthogonal images account for every primitive idempotent in the target, because what remains is nilpotent. Thus primitive idempotents under a surjection of commutative Artinian ideals apply: the primitive idempotents of $J_D$ correspond exactly to primitive central idempotents $b\in I_D$ with ${\mathrm{Br}}_D^G(b)\ne 0$.

To identify these with the required blocks, we prove the trace criterion for defect groups. Choose a subgroup $D_0$ of smallest order with $b\in I_{D_0}$. Such a subgroup exists: for a Sylow $p$-subgroup $P$, $b={\mathrm{Tr}}_P^G(b/[G:P])$. If ${\mathrm{Br}}_{D_0}^G(b)=0$, write $b={\mathrm{Tr}}_{D_0}^G(a)$ and replace $a$ by $ba$. The kernel formula and transitivity of trace would giveMultiplying by $b$ gives the identity of the local algebra $bZ(A)$ in a sum of the ideals $b{I}_Q$. If all were proper, they would lie in its unique maximal ideal; hence some $b{I}_Q$ contains $b$, contradicting minimality. Therefore ${\mathrm{Br}}_{D_0}^G(b)\ne 0$.

The general trace-vanishing assertion shows that any $E$ with nonzero Brauer image is conjugate into $D_0$. Conversely, every conjugate of $D_0$ has nonzero image. Hence the maximal such subgroups, each called a defect group of a block, are precisely the conjugates of $D_0$. It also follows, using trace transitivity, that $b\in I_D$ if and only if a defect group is conjugate into $D$. Combining this with ${\mathrm{Br}}_D^G(b)\ne 0$ forces equality of the subgroup orders, so the idempotents singled out above are exactly those with defect group $D$.

We have proved the desired bijectionFinally, Brauer first main theorem states that these blocks correspond bijectively to the blocks of $k{N}_G(D)$ with defect group $D$. Apply the same argument to $N$: its normalizer of $D$ is itself and its centralizer of $D$ is still $C$. The target ideal is therefore the same $J_D$. Matching the two bijections proves the theorem and gives the Brauer correspondence, characterized by equal nonzero Brauer images. Notice that we never assume the entire kernel on $I_D$ is nilpotent; it may kill whole blocks of smaller defect.

The decomposition matrix separates into two connected components. The first contains ${\varphi }_1,{\varphi }_2$ and ${\chi }_1,{\chi }_{3A},{\chi }_{3B},{\chi }_4$; it is the principal block $B_0$. The second contains only ${\varphi }_3$ and ${\chi }_5$; since ${\varphi }_3(1)=5$ contains the full 5-part of $|A_5|=60$, this is a defect-zero representation and its block $B_1$ has defect group $1$.

For $D=1$, the normalizer is $N_G(1)=G$, so the Brauer correspondence is the identity and $B_1$ corresponds to itself.

The defect group of the principal block is a Sylow 5-subgroup $P\cong C_5$. There are six Sylow 5-subgroups in $A_5$, so the orbit-stabilizer theorem givesThe centralizer of a 5-cycle in $A_5$ is $P$, and an involution in the normalizer acts on $P$ by inversion. HenceIn characteristic five the simple $kN$-modules are inflated from $N/P\cong C_2$: their Brauer characters are ${\psi }_{+}=(1,1)$ and ${\psi }_{-}=(1,-1)$ on the identity and involution classes. If $1,\epsilon ,{\rho }_1,{\rho }_2$ are the two one-dimensional and two two-dimensional ordinary characters of $D_{10}$, their reductions areThe resulting decomposition matrix is connected, so these characters form the unique 5-block $b_0$ of $N$, with defect group $P$. By the Brauer first main theorem,This is the complete 5-modular blocks of A5 correspondence.