Put $N_G=\sum_{g\in G}g$. Its image on every module lies in the <invariant submodule>, since $hN_G=N_G$. In the regular module,
$$
(kG)^G=kN_G,
$$
and this line is the socle of the projective cover $P_k$ of the trivial module.

Suppose $N_GM\ne0$. Choose $m\in M$ with $N_Gm\ne0$ and consider the homomorphism
$$
f:kG\longrightarrow M,
\qquad x\longmapsto xm.
$$
Its restriction $f|_{P_k}$ is nonzero on $\operatorname{Soc}(P_k)=kN_G$. Because $P_k$ is the <injective hull> of its simple socle, that socle is essential: every nonzero submodule meets it. Hence $\ker(f|_{P_k})=0$. The resulting embedding $P_k\hookrightarrow M$ splits because $P_k$ is injective. Since $M$ is indecomposable, $M\cong P_k$.

Conversely, on $P_k$ the image of $N_G$ is its one-dimensional socle. Thus the <group norm element detects the trivial projective cover>:
$$
\boxed{
\dim_k(N_GM)=
\begin{cases}
1,&M\cong P_k,\\
0,&M\not\cong P_k.
\end{cases}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 3 / d / Solution

{title2=$N_G=\sum_{g\in G}g$}

Let $M$ be an indecomposable finite-dimensional $kG$-module and let $P_k$ be the projective cover of the trivial module. Then
$$
\dim_k(N_GM)=
\begin{cases}1&M\cong P_k,\\0&\text{otherwise}.
\end{cases}
$$


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