The Brauer table of S3 in characteristic two

Set $r=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ and $t=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Direct multiplication gives $r^2=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, $r^3=t^2=I$, and $trt=r^2$. The six matrices $I,r,r^2,t,rt,r^{2}t$ are distinct. Acting on the three nonzero vectors of ${\mathbb{F}}_2^2$ embeds this group in $S_3$; its size six identifies it with $S_3$.

The polynomial of r is $X^2+X+1$, with distinct nontrivial cube roots $\lambda ,{\lambda }^2$ in k. An invariant line for r must be one of its eigenlines. From $rt=t{r}^{-1}$, t sends the $\lambda$-eigenline to the ${\lambda }^{-1}$-eigenline, which is different. Hence no line is invariant under both; the two-dimensional representation is simple, and the same argument holds over every extension field.

Writing $\zeta =\widehat{\lambda }$, the identity has value 2 and r has value $\zeta +{\zeta }^2=-1$, since ${\zeta }^3=1$ and $\zeta \mathrm{\ne }1$. The trivial representation has values 1,1. Thus the determinant is $-1-2=-3$, nonzero in K and with residue 1 in k.

The p-regular permutations are the identity and the two conjugate 3-cycles, hence give exactly two classes. By independence at most two nonisomorphic simple modules can have characters on this two-dimensional space of class functions. The two already exhibited are nonisomorphic by dimension and simple, so exhaust the possibilities.