For which characteristic is the standard representation of $S_n$ semi-simple? Let $k$ be a field with characteristic $p$. Let $S_n$ be the symmetric group. By maschke's theorem we know that if $p \nmid n!$ then any $kS_n$-module is semi-simple. However if we look at the standard representation of $S_3$ we have a stronger result. Namely it is semi-simple if and only if $p \nmid 3$. Are there similar results for arbitrary $n$? 
 A: Unfortunately the "standard" representation sometimes refers to $k^n$ with the action permuting the coordinates, and some times it refers to the $(n-1)$ dimensional subspace where the sum of the coordinates is zero. Obviously these two representations are closely related, but it is a bit unfortunate that the word "standard" has this problem. Anyway I digress.
In addition to this $(n-1)$-dimensional subspace, $k^n$ also has a one-dimensional invariant subspace where all the coordinates are equal.  Here is where the funny business happens though: Normally these two spaces are complements to one another, but if $p$ divides $n$ then the invariant subspace is contained in the $(n-1)$ dimensional subspace and neither of them admits an invariant complement.
In this case we can take the $(n-1)$ dimensional subspace and quotient it by the $1$ dimensional space of invariants to obtain a $(n-2)$ dimensional irreducible representation which is not the reduction mod p of a complex representation.
A: Let $k$ be a field and $p = \operatorname{char} k$.
For $G = S_n$ we have the standard permutation action on $X = \{1, \ldots, n\}$, which gives us a permutation module $kX$, where $X$ is the basis of $kX$ and $G$ acts by permutation matrices.
You can see that $kX$ has a unique $1$-dimensional submodule , namely the one spanned by $$z = \sum_{ x \in X } x.$$
(You should prove this).
There is also a submodule $W$ of codimension $1$, namely $$W = \left\{\sum_{x \in X} \lambda_x x : \sum_{x \in X} \lambda_x = 0\right\}.$$
If $p \mid n$, then $z \in W$. Since $z$ spans the unique $1$-dimensional submodule of $kX$, it follows that $W$ has no complement in $kX$. Thus $kX$ is not semisimple.
If $p \nmid n$, then $W$ is irreducible (prove this) and $kX = W \oplus \langle z \rangle$, so $kX$ is semisimple.
