Indecomposable Submodule of Group algebra Suppose $k$ is a field with characteristic $3$ and let $G = S_3$. Consider the group algebra $A = kG$, and let $e=1/2(1+(12))$. Show that any submodule $N\leq Ae$ contains the submodule $$M = \{m\in Ae : (123)m = m\},$$ and hence conclude that $Ae$ is indecomposable.
I'm having trouble with one aspect of this question. I know how to conclude that $Ae$ is indecomposable, because if every submodule contains $M$ then it's impossible to write $M$ as a direct sum since then we would need two submodules with trivial intersection.
I can also conclude that if $N$ has non-trivial intersection with $M$, then $N$ contains all of $M$, since $M$ is one-dimensional.
My issue is showing that $N\cap M\neq 0$. Since $N$ is arbitrary, I can't see how to approach the question.
Any help would be appreciated.
 A: Suppose $N\leq Ae$ and let $x\in N$ be non-zero. Set $g=(123)$, and note that $\langle g\rangle \cong \mathbb{Z}/3$.
Note that as a $k[\langle g\rangle]\cong k[\mathbb{Z}/3]$-module, $Ae \cong k[\mathbb{Z}/3]$ is free of rank 1. I claim that 
$$ M = (Ae)^{\mathbb{Z}/3} = Nm\cdot Ae,$$
 where $Nm = 1+(123)+(132)=1+g+g^2$ is the norm element of $k[\langle (123)\rangle]\cong k[\mathbb{Z}/3]$. To see this, let $a+bg+cg^2\in k[\mathbb{Z}/3]^{\mathbb{Z}/3}$ be an invariant element. Then $$g\cdot(a+bg+cg^2)=c+ag+bg^2=a+bg+cg^2,$$ forcing $a=b=c$, so that $a+bg+cg^2 = a\cdot Nm$ is a multiple of the norm. The other inclusion is easy.
If $x\in M$, then we are done as $M\cap N\neq 0$. Otherwise, $Nm\cdot x\in N$ is invariant under $g$. Provided it is non-zero, we also get $M\cap N\neq0$. It vanishes only if $$x\in \ker(Nm\cdot) = (1-g)Ae,$$ where the equality comes from another simple computation in $k[\mathbb{Z}/3]$.Therefore, if $Nm\cdot x =0$, then set $x=(1-g)x'$.
Note that because $\mathrm{char}(k)=3$, $(1-g)x = Nm\cdot x'\in N$ is invariant if non-zero. But $$\ker(1-g) = Nm\cdot Ae=M,$$ so $(1-g)x\neq0$ by our assumption that $x$ is not invariant (ie: $x\notin M$). Thus $(1-g)x\in N$ is an invariant element, forcing $M\cap N\neq 0$. 
Thus, $M\cap N\neq 0$ in all cases, and your argument finishes things off. 
