Is my following proof correct?
Let $G$ be a finite group and $\mathbb{C}G$ its group algebra. Let $\phi:\mathbb{C}G\to\mathbb{C}G$ be a $\mathbb{C}G$-homomorphism.
$ i)$ Then there exists a $w\in\mathbb{C}G$ such that $\phi(r)=rw$.
$ii)$ Let $W\subset\mathbb{C}G$ be an irreducible submodule and choose $0\neq w\in W$. Then $W=\{rw\colon r\in\mathbb{C}G\}$.
Proof of $i)$. Let $e$ be the neutral element in the group, define $w:=\phi(e)$. Then since $\phi$ is a $\mathbb{C}G$-homomorphism, we see that $$ \phi(r)=\phi(r\cdot e)=r\cdot\phi(e)=rw. $$
Proof of $\textit{ii)}$. I want to prove that $\{rw\colon r\in\mathbb{C}G\}$ is a submodule of $W$. Since it is non-zero and $W$ is irreducible, it then follows that $W=\{rw\colon r\in\mathbb{C}G\}$. For showing it is a submodule, I wanted to use that $gW\subset W$ for all $g\in G$. Then, is it true that $rW\subset W$ for all $r\in\mathbb{C}G$ and why?
$W'=\{rw\mid r\in\mathbb{C}G\}$ is a $\mathbb{C}G$-submodule: for any $w_1,w_2\in W'$, $r\in\mathbb{C}G$, we have $w_1=r_1w$ and $w_2=r_2w$ for some $r_1,r_2\in\mathbb{C}G$. Then $$ rw_1+w_2=rr_1w+r_2w=\underbrace{(rr_1+r_2)}_{\in\mathbb{C}G}w\in W'. $$