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For $N=N_G(D)$ and $C=C_G(D)$, the Brauer morphism restricts to a surjectionThis follows from Brauer morphism and relative trace and the fact that $kC$ is contained in $(kG{)}^D$. The target is an ideal in the commutative algebra $(kC{)}^N$. Applying primitive idempotents under a surjection of commutative Artinian ideals and the trace criterion for defect groups identifies block idempotents with defect group $D$ with the primitive idempotents of $J_D$. Applying the same argument to $N$ gives Brauer first main theorem. The whole kernel need not be nilpotent, since blocks with smaller defect can be killed.

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.