$G$ abelian if $\Leftrightarrow$ all its irreducible representations are of degree $1$ 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!
 A: *

*This shows that $\rho_g$ is linear because $(\rho_h \circ \rho_g)(v)
    = h \cdot \rho_g(v)$, and $(\rho_g \circ \rho_h)(v) = \rho_g(h \cdot v)$ so this shows that $\rho_g$ is $G-$linear.

*I'm assuming there was a hypothesis that $V$ is a finite dimensional representation, or otherwise your text is only working with finite dimensional representations. For a finite group, all the irreducible representations are finite dimensional (there are several proofs of this fact you can find easily), and so it's also possible this theorem will only be used on irreducible representations of finite groups.

*Yep, it's because $\rho_g = \lambda \operatorname{Id}$. In particular, the set $\operatorname{span}(e_1)$ is the set $\{k e_1 : k \in \mathbb{C}\}$. Now, if $c \in \mathbb{C}$, taking apply $\rho_g(c e_1) = c\rho_g(e_1) = c\lambda e_1$. Then, for any $k$, setting $c = \frac{k}{\lambda}$, we get that $\rho_g(c e_1) = k e_1$, and so $\rho_g(\operatorname{span}(e_1)) = \operatorname{span}(e_1)$. More concisely, $\rho_g(\operatorname{span}(e_1))$ is a one dimensional subspace which contains $e_1$, and so it must be $\operatorname{span}(e_1)$

*I'm assuming you mean that $V_{i} = \operatorname{span}(e_i)$. We are allowed to do this because, by assumption, the irreducible representations are all one dimensional. Now, we want to write $\rho_g$ as a matrix; we do this by checking what it does to a basis of $V$. The set $\{e_i\}_i$ is a basis, since $V = \bigoplus_{i}V_i$, and so every element of $V$ can uniquely be written as a linear combination of the $e_i$, which is equivalent to the $e_i$ being a basis. Then, because the irreducible representations are each themselves invariant under $\rho_g$ for any $g$. This means that $\rho_g(e_i) = \lambda_ie_i$ for some $\lambda_i \in \mathbb{C}$. Therefore, $\rho_g$ is a diagonal matrix, with $\lambda_i$ on the diagonals.

*How does this complete the proof? We know that diagonal matrices commmute with each other, and we know there is some injective homomorphism $\rho : G \to GL(V)$ (for some $V$). This means the image of $G$ is abelian, and since the map is injective, this implies that $G$ is abelian.

I hope this helps!
