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Put $a=(123)$ and $t=(12)$. Conjugation by $a$ fixes the three given idempotents of $k⟨a⟩$, while $ta{t}^{-1}=a^{-1}$ fixes $e_1$ and interchanges $e_2,e_3$. Their orbits are therefore $\boxed{\{e_1\},\{e_2,e_3\}}$, with central orbit sumsusing $\omega +{\omega }^2=1$ in characteristic $2$.

To ensure these sums really are primitive central idempotents, examine the two ideals. The first has basis $b_0,b_{0}t$, with $b_{0}a=b_0$, so $b_{0}kG\cong k{C}_2$, a local ring. The second has dimension $4$: $b_{1}k⟨a⟩=k{e}_2\oplus k{e}_3$ has dimension $2$, and the two cosets of $⟨a⟩$ double it. In the two-dimensional representationThese matrices satisfy $a^3=t^2=1$ and $ta{t}^{-1}=a^{-1}$, so they define a representation of $S_3$. In it, $b_1$ acts as identity and $b_0$ as zero. The distinct diagonal entries supply the two diagonal matrix units, and multiplying by the swap matrix supplies the off-diagonal units. The induced map $b_{1}kG\to M_2(k)$ is thus surjective and, by dimension, an isomorphism. Both summands have no nontrivial central idempotents, so $b_0,b_1$ are exactly the 2-modular blocks of S3.

The Brauer morphism at the trivial subgroup is the identity. For $H=⟨t⟩$, $C_G(H)=H$, and neither $a$ nor $a^2$ lies in this centralizer. ThusIndeed $H$ is Sylow, and all nontrivial $2$-subgroups are conjugate to it; these images therefore determine maximality in the definition of a defect group of a block. The original PDF provides the $S_3,{\mathbb{F}}_4,N,H$ setup missing from the TeX transcription.

Regard the block algebra $RGe$ as an $R[G\times G]$-module through left and right multiplication,A defect group of a block is a p-subgroup $D\le G$ for which $\Delta D=\{(d,d):d\in D\}$ is a vertex of an indecomposable summand determining the block; equivalently, $D$ is maximal withunder the Brauer morphism. The uniqueness of vertices up to conjugacy in $G\times G$, together with the diagonal form of these vertices, shows that any two such $D$ are conjugate in $G$. Hence

A p-subgroup is p-radical if it equals the p-core of its normalizer. Every defect group of a block is p-radical by defect groups are centralizer-conjugate Sylow intersections. Being an intersection of arbitrary Sylow subgroups alone is weaker than the centralizer-conjugate statement used in that argument.

For a $p$-subgroup $D$ of a finite group $G$, set $I_D={\mathrm{Tr}}_D^G((kG{)}^D)$, a transfer ideal of conjugation-fixed elements in $Z(kG)$. A block of a group algebra idempotent $b$ lies in $I_D$ if and only if a defect group of a block of $b$ is conjugate to a subgroup of $D$. In particular, defect groups can equivalently be defined as minimal subgroups $D$ with $b\in I_D$.

Here is a proof using the Brauer morphism. A subgroup of smallest order with $b\in I_D$ exists, since a Sylow subgroup $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_D^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_D(b)=0$, then $ba$ belongs to the kernel of the Brauer morphism, which is the sum of proper-subgroup relative traces. Hence $b\in {\sum }_{Q<D}{I}_Q$. The algebra $bZ(kG)$ is a commutative Artinian ring and a local ring, so multiplying this equality by $b$ forces some ideal $b{I}_Q$ to contain its identity $b$. This contradicts minimality. On the other hand, ${\mathrm{Br}}_E(I_D)=0$ unless $E$ is conjugate into $D$: the $E$-action on $G/D$ has no fixed coset otherwise, and nonfixed orbit sizes vanish in characteristic $p$. Thus these minimal subgroups are precisely the maximal subgroups with nonzero Brauer image. They are all conjugate. Trace transitivity proves the stated criterion for larger $D$.