Let $G=S_r$ , the permutation group on $r$ elements, and $N=k^r$ where $k$ is an arbitrary field. Consider the permutation representation of $G$ on $N$ given by$$\sigma(a_1,\ldots,a_r)=(a_{\sigma(1)},\ldots,a_{\sigma(r)}),\ \sigma\in S_r, a_i\in $$

If $r>1$ , we can define two submodules of $N$ , called the trace and augmentation, as

If the characteristic of $k$ divides $r$ , then obviously $N''\supset N'$ . Otherwise, $N''$ is a simple (irreducible) $G$ -module. For suppose $N''$ has a nontrivial submodule $M$ , and choose a nonzero $u\in M$ . Then some pair of coordinates of $u$ are unequal, for if not, then $u=(a,\ldots,a)$ and then $u\not\in N''$ because of the restriction on the characteristic of $k$ forces $ra\neq 0$ . So apply a suitable element of $G$ to get another element of $M$ , $v=(b_1,b_2,\ldots,b_r)$ where $b_1\neq b_2$ (note here that we use the fact that $M$ is a submodule and thus is stable under the action of $G$ ).

But now $(1 2)v - ev = (b_1-b_2,b_2-b_1,0,\ldots,0)$ is also in $M$ , so $w=(1,-1,0,\ldots,0)\in M$ . It is obvious that by multiplying $w$ by elements of $k$ and by permuting, we can obtain any element of $N''$ and thus $M=N''$ . Thus $N''$ is simple.

It is also obvious that $N=N'\oplus N''$ .