= Solution
{c}
Write $A=kG$, $N=N_G(D)$ and $C=C_G(D)$. The <Brauer morphism> is the coefficient projection
$$
\operatorname{Br}_D^G:A^D\longrightarrow kC,\qquad\sum_{g\in G}a_gg\longmapsto\sum_{g\in C}a_gg,
$$
where 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 is
$$
\ker\operatorname{Br}_D^G=\sum_{Q<D}\operatorname{Tr}_Q^D(A^Q).
$$
Indeed, 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> gives
$$
\operatorname{Br}_D^G\bigl(\operatorname{Tr}_D^G(a)\bigr)=\operatorname{Tr}_D^N\bigl(\operatorname{Br}_D^G(a)\bigr).
$$
One 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, $\operatorname{Br}_E^G(\operatorname{Tr}_D^G(A^D))=0$ unless $E$ is conjugate into $D$.
Set $I_D=\operatorname{Tr}_D^G(A^D)$ and $J_D=\operatorname{Tr}_D^N(kC)=(kC)_D^N$. 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 homomorphism
$$
\operatorname{Br}_D^G:I_D\twoheadrightarrow J_D.
$$
The 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 $\operatorname{Br}_D^G(b)\ne0$.
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 $P$, $b=\operatorname{Tr}_P^G(b/[G:P])$. If $\operatorname{Br}_{D_0}^G(b)=0$, write $b=\operatorname{Tr}_{D_0}^G(a)$ and replace $a$ by $ba$. The kernel formula and transitivity of trace would give
$$
b\in\sum_{Q<D_0}I_Q.
$$
Multiplying by $b$ gives the identity of the local algebra $bZ(A)$ in a sum of the ideals $bI_Q$. If all were proper, they would lie in its unique maximal ideal; hence some $bI_Q$ contains $b$, contradicting minimality. Therefore $\operatorname{Br}_{D_0}^G(b)\ne0$.
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 $\operatorname{Br}_D^G(b)\ne0$ forces equality of the subgroup orders, so the idempotents singled out above are exactly those with defect group $D$.
We have proved the desired bijection
$$
\boxed{b\longmapsto\operatorname{Br}_D^G(b):\{\text{blocks of }kG\text{ with defect }D\}\overset{\sim}{\longrightarrow}\operatorname{Prim}(J_D)}.
$$
Finally, <Brauer first main theorem> states that these blocks correspond bijectively to the blocks of $kN_G(D)$ 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.
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