= Solution
{c}
Put $a=(123)$ and $t=(12)$. Conjugation by $a$ fixes the three given idempotents of $k\langle a\rangle$, while $tat^{-1}=a^{-1}$ fixes $e_1$ and interchanges $e_2,e_3$. Their orbits are therefore $\boxed{\{e_1\},\{e_2,e_3\}}$, with central orbit sums
$$
b_0=e_1=1+a+a^2,\qquad b_1=e_2+e_3=a+a^2,
$$
using $\omega+\omega^2=1$ in characteristic $2$.
To ensure these sums really are primitive central idempotents, examine the two ideals. The first has basis $b_0,b_0t$, with $b_0a=b_0$, so $b_0kG\cong kC_2$, a <local ring>. The second has dimension $4$: $b_1k\langle a\rangle=ke_2\oplus ke_3$ has dimension $2$, and the two cosets of $\langle a\rangle$ double it. In the two-dimensional representation
$$
a\longmapsto\begin{pmatrix}\omega&0\\0&\omega^2\end{pmatrix},\qquad t\longmapsto\begin{pmatrix}0&1\\1&0\end{pmatrix},
$$
These matrices satisfy $a^3=t^2=1$ and $tat^{-1}=a^{-1}$, 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 $b_1kG\to M_2(k)$ 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 $H=\langle t\rangle$, $C_G(H)=H$, and neither $a$ nor $a^2$ lies in this centralizer. Thus
$$
\boxed{\begin{array}{c|c|c|c}
\text{block}&\operatorname{Br}_1^G&\operatorname{Br}_H^G&\text{defect group}\\\hline
b_0=1+a+a^2&b_0&1&H\\
b_1=a+a^2&b_1&0&1
\end{array}}.
$$
Indeed $H$ 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.
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