Understanding the proof that the trivial $\mathbb{C}G$-submodule of $\mathbb{C}G$ is unique. Question:
Let $G$ be a finite group. Find a $\mathbb{C}G$-submodule of $\mathbb{C}G$ which is isomorphic to the trivial $\mathbb{C}G$-module. Is there only one such $\mathbb{C}G$-submodule?
Answer:
Let $V = sp( \sum_{g \in G}  g ) $. Then V is a trivial $\mathbb{C}G$-submodule of $\mathbb{C}G$. Now suppose that $U$ is an arbitrary trivial $\mathbb{C}G$-submodule of $\mathbb{C}G$, so $U = sp(u)$ for some $u$. Then $gu=u$ for all $g \in G$, so $u|G| = (\sum_{g\in G}g)u = u(\sum_{g\in G}g) \in V$. Thus $U=V$, and so $\mathbb{C}G$ has exactly one trivial $\mathbb{C}G$-submodule, namely $V$.
My questions:
What is the definition of a trivial submodule?
Why does $U$ being a trivial submodule mean it is generated by a single generator, $u$?
Why does $(\sum_{g\in G}g)u = u(\sum_{g\in G}g)$?
 A: In this context, a trivial submodule of $\mathbb{C}G$ is a simple submodule on which elements of $G$ act trivially. That is, if $V$ is a trivial submodule and $v\in V$ then $gv=v$.
Now, given this definition, it is clear that any trivial submodule must be one dimensional. Indeed, if $U\subset\mathbb{C}G$ is a trivial submodule and $u\in U$ is nonzero, then the 1-dimensional subspace $\mathbb{C}u$ is a proper $G$-submodule (since $gu=u$ for all $g\in G$). Hence $\mathbb{C}u=U$.
Now, write $u=\sum_{h\in G}\lambda_hh$ for some $\lambda_h\in \mathbb{C}$. To see that $u\left(\sum_g g\right)=\left(\sum_g g\right)u$ it is enough to observe that
$$h\left(\sum_{g\in G} g\right)=\left(\sum_{g\in G} g\right)h$$
for every $h\in G$. To this end, note that
$$h\left(\sum_{g\in G} g\right)=\left(\sum_{g\in G} hg\right)=\left(\sum_{g\in G} (hgh^{-1})h\right)=\left(\sum_{g\in G} (hgh^{-1})\right)h.$$
However, the map $g\mapsto hgh^{-1}$ is a bijection $G\to G$, so
$$\sum_{g\in G} (hgh^{-1})=\sum_{g\in G}g$$
as required.
