example of reducible and irreducible $G$-modules

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

$$\sigma (a_1,\mathrm{\dots },a_r)=(a_{\sigma (1)},\mathrm{\dots },a_{\sigma (r)}),\sigma \in S_r,a_i\in k$$

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

$$N^{\prime }=\{(a,a,\mathrm{\dots },a)\}$$

$$N^{\prime \prime }=\{(a_1,a_2,\mathrm{\dots },a_r)|\sum a_i=0.\}$$

Clearly both $N^{\prime }$ and $N^{\prime \prime }$ are stable under the action of $G$ and thus in fact form submodules of $N$.

If the characteristic of $k$ divides $r$, then obviously $N^{\prime \prime }\supset N^{\prime }$. Otherwise, $N^{\prime \prime }$ is a simple (irreducible) $G$-module. For suppose $N^{\prime \prime }$ 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 $u=(a,\mathrm{\dots },a)$ and then $u\notin N^{\prime \prime }$ because of the restriction on the characteristic of $k$ forces $ra\ne 0$. So apply a suitable element of $G$ to get another element of $M$, $v=(b_1,b_2,\mathrm{\dots },b_r)$ where $b_1\ne b_2$ (note here that we use the fact that $M$ is a submodule and thus is stable under the action of $G$).

But now $(12)v-ev=(b_1-b_2,b_2-b_1,0,\mathrm{\dots },0)$ is also in $M$, so $w=(1,-1,0,\mathrm{\dots },0)\in M$. It is obvious that by multiplying $w$ by elements of $k$ and by permuting, we can obtain any element of $N^{\prime \prime }$ and thus $M=N^{\prime \prime }$. Thus $N^{\prime \prime }$ is simple.

It is also obvious that $N=N^{\prime }\oplus N^{\prime \prime }$.