Blocks and defect groups of s3

In characteristic $2$, put $f=a+a^2$ and $e=1+f$. Since $a^3=1$, one has $f^2=a^2+a=f$ and $e^2=e$, with $ef=0$ and $e+f=1$. A transposition $t$ conjugates $a$ to $a^{-1}$, so both idempotents are central. Also $ae=e$, and $ekG$ has basis $e,et$, with $(et{)}^2=e$. Thus it is $k{C}_2\cong k[u]\mathrm{/}((u-1{)}^2)$. An element is a unit exactly when its constant term in $u-1$ is nonzero, by a two-term geometric inverse. This local algebra has no nontrivial idempotent, so $e$ is primitive central.

Define matrices $A=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ and $T=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ over $k$. Direct multiplication gives $A^3=T^2=I$ and $TAT=A^{-1}$. These are the relations of $S_3=⟨a,t\mid a^3=t^2=1,tat=a^{-1}⟩$: the relations reduce every word to one of $a^i{t}^j$ with $0\le i<3$, $0\le j<2$, and the actual permutations give six distinct forms. Hence $a\mapsto A,t\mapsto T$ defines a representation and its linear extension $\rho :kG\to M_2(k)$ is an algebra map. One has $\rho (f)=A+A^2=I$. The four matrices $I,A,T,AT$ span $M_2(k)$: the standard matrix units are $E_{22}=A+T$, $E_{11}=I+A+T$, $E_{21}=AT+I$, and $E_{12}=T+AT+I$. They are the images of $f,fa,ft,fat$. These four group-algebra elements form a basis of $fkG$, since $f,fa$ are independent in $fk⟨a⟩$, span that ideal by $f{a}^2=f+fa$, and the two cosets of $⟨a⟩$ have disjoint supports. Thus the restriction of $\rho$ is a surjective map between four-dimensional algebras, hence an isomorphism $fkG\cong M_2(k)$. Its centre is $kf$, so $f$ is primitive and $e,f$ are all blocks. The augmentation of $e$ is $3=1$, so $e$ acts as identity on the trivial module and is principal. This calculation uses no root of unity in $k$.

For $P=⟨t⟩$, direct commutation in $S_3$ gives $C_G(P)=P$. Neither $a$ nor $a^2$ centralizes $t$, hence ${\mathrm{Br}}_P(e)=1$ and ${\mathrm{Br}}_P(f)=0$. All nontrivial $2$-subgroups are transposition subgroups. At $1$, the image of $f$ is $f\mathrm{\ne }0$. Thus [F3] gives Sylow defect for $e$ and defect $1$ for $f$, agreeing with [F1] for the principal block and [F2] for the explicitly computed matrix block.

In characteristic $3$, let $T$ be the sum of the three transpositions and $C=a+a^2$. The centre has basis $1,T,C$, because commuting with all group elements is equivalent to having constant coefficients on conjugacy classes. Direct multiplication gives $T^2=3(1+C)=0$, $C^2=C+2$, and $TC=2T$. Thus $J=kT+k(1+C)$ satisfies $J^2=0$, and the centre is $k1\oplus J$. For $z=\lambda 1+u$ with $u\in J$, the equation $z^2=z$ is ${\lambda }^2=\lambda$ and $(2\lambda -1)u=0$. The first equation gives $\lambda =0$ or $1$; in both cases $2\lambda -1\mathrm{\ne }0$ in characteristic $3$, so $u=0$. Therefore the only central idempotents are $0,1$. There is one block, necessarily principal; [F1] gives defect $⟨a⟩\cong C_3$.