I'm new to the group representation theory and I'm trying to understand the proof of
$G$ finite group.
$G$ abelian $\Leftrightarrow$ all its irreducible representations are of degree $1$.
Let $\rho$ be a irreducible representation of $G$ in $V$. Take $g \in G$, $\forall (h,v) \in G \times V$, we have $\rho_g \circ \rho_h(v) = \rho_h \circ \rho_g(v)$, because $G$ abelian. This shows that $\rho_g$ is $G$-linear. (Why?)
By the Schur lemma, $\rho_g = \lambda \text{Id}$ where $\lambda \in \mathbb{C}$.
Now suppose that $V = \text{span}(e_1,..,e_n)$. (Why can we suppose that? Is it because $G$ is finite?)
$\text{span}(e_1)$ is a vectorial space of $V$, stable by $\rho_g$ (Why? I suppose that it is because we showed $\rho_g = \lambda \text{Id}$, but can't see why exactly).
So $\text{span}(e_1)$ is a sub-$\mathbb{C[G]}$-module non zero of $V$. Finally $V = \text{span}(e_1)$.
Let $\rho$ be a representation of $G$ in $V$. We write $V_i = V_1 \oplus V_2 \dots \oplus V_k$ and $V = \text{span}(e_i)$. $\forall g \in G, \exists (\lambda_i) \in \mathbb{C}^n$ such that $\rho_g$ is a diagonal matrix with $\lambda_i$ on its diagonal. , thus $\rho(G)$ is commutative because orthogonal matrices are commutative.
This whole last part is also not clear for me. I hope there is someone who could explain the different steps to me so that I understand the proof well. Thanks!
Best Answer
I hope this helps!