Let $G=\mathfrak{S}_3=\langle (12),(23)\rangle$, the symmetric group on $3$ letters, $k$ a field and $V$ the "standard module", $V=\langle v_1, v_2, v_3 \rangle$ and the group algebra $A=k[G]$.
If the characteristic of $k$ doesn't divide $|G|$, we can write $V=V_1 \oplus V_2$ with $V_1=\langle v_1+v_2+v_3 \rangle$ and $V_2=\langle (v_1-v_2), (v_2-v_3) \rangle$ i.e. $\{v_1-v_2, v_2-v_3\}$ is a basis for the orthogonal complement $V_1$ of $V$.
$\mathfrak{S}_3$ acts on $k^3$ by permuting the coordinates of a column vector: $\rho_g(v_i)=v_{g(v_i)}$.
$V_1$ is a $\mathfrak{S}_3$-invariant subspace of $V$; its complement $V_1^{\perp}=V_2$ can be defined as the kernel of the projection of $V$ onto $V_1$: \begin{equation} V_2=\{(a, b, c) \}\: | \: a+b+c=0 \}. \end{equation}
If characteristic of $k$ is $3$, then we need to find a basis of $V$ to exhibit the following chain called a $\textbf{flag}$ (fixed by $G$):
\begin{equation} \displaystyle V_1=\langle v_1+v_2+v_3 \rangle \subset V_2=\langle v_1+v_2+v_3, v_1-v_2 \rangle \subset V \end{equation}
Seeing this chain of inclusions, we can think of a basis for $V$ written as $\{v_1+v_2+v_3, v_1-v_2, \bullet \bullet \bullet \}$ but I can't see why this particular basis of $V$ works.
Hope my question makes sense ! I'm not really sure about that ! Positive characteristic is a complicated subject. I thank you in advance for any suggestions.