Proving complete reducibility of modular representations Let $G$ = $S_{3}$ and consider the $3 \times 3 $ permutation representations. For example, we have 
$$
\psi (123) = 
\begin{pmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{pmatrix}.
$$ How would one show that this representation is completely reducible over characteristic $2$ but not completely reducible over characteristic $3$?
 A: Any permutation representation in $V=K^n$ has an invariant subspace $W$ given by the equation $x_1+x_2+\cdots+x_n=0$ (the vectors with coordinates summing to$~0$; here of course $n=3$). Since $\dim W=n-1$, any complementary subspace to $W$ is spanned by some vector $v\notin W$. In order for that complementary subspace to be invariant, $v$ must be an eigenvector for the action of any element of the group: its span must define a $1$-dimensional representation of the group. Now the symmetric group $S_3$ has no other $1$-dimensional representations than the trivial representation and the sign representation (this is true in any characteristic); in both cases a $3$-cycle acts as the identity. So $v$ must be invariant under cyclically permuting its coordinates, which means all coordinates of $v$ must be equal.
Now if $K$ does not have characteristic$~3$ one can take $v=(1,1,1)$, which spans a complementary invariant subspace to$~W$ (and in fact the representation is completely reducible, since $W$ is irreducible). However in characteristic$~3$ one has $(1,1,1)\in W$ (and indeed $(x,x,x)\in W$ for any $x\in K$), so $W$ does not admit a complementary invariant subspace; the representation is not completely reducible.
A: For $G$ a finite group, and $F$ is a field, the group algebra $FG$ is semisimple if and only if $J(FG) = \{0\}$, where $J(FG)$ is the Jacobson radical off $FG$, which has several equivalent definitions: one is $\{ j \in FG: 1-jx \}$ is nilpotent for all $x \in FG$.
We note that if $j \in Z(FG)$ is nilpotent and non-zero, then $jx$ is nilpotent for all $ x \in FG,$ so $J(FG) \neq \{0\}.$ Now when $F$ has characteristic $p \neq 0$ and $G$ has a non-identity normal $p$-subgroup $U$, then it is know $1-u \in J(FG)$ for all $u \in U.$
This is the case when $G = S_{3}$ and $F$ has characteristic $3,$ since $U = A_{3}$ will do.
Another characterization of $J(FG)$ is that it is the unique largest ideal which annihilates  every irreducible $FG$ module. Therefore, any $FG$-module $M$ which is not annihilated by $J(FG)$ can't be completely reducible. Now $(123) - 1_{S_{3}} \in J(FS_{3})$, where $F$ has characteristic $3,$ but does not annihilate the given module.
In fact, we can see (half of) Maschke's theorem from this point of view: if ${\rm char}(F)$ divides $|G|$, then $G^{+} = \sum_{g \in G} g$ in $FG$ is nilpotent, non-zero, and in $Z(FG)$, so $FG$ is not semisimple.
