All linear representations of a finite group are quiver representations? Definitions


*

*A linear representation of a finite group $G$ is a group homomorphism:
$$\rho:G\to\text{GL}(V),$$
where we shall assume that $V$ is a finite dimensional vector space. We call $V$ the representation space of $\rho$.

*Let $Q$ be a finite quiver with (finite) vertex set $Q_0$ and (finite) edge set $Q_1$. Then a representation $M,$ of the quiver $Q$, is given by assigning 1) for each $i\in Q_0$, a $k$-vector space $M_i$. 2) for each $\alpha\in Q_1$, a $k$-linear homomorphism $\varphi_\alpha:M_{s(\alpha)}\to M_{t(\alpha)}$.


(We write $s(\alpha)$ and $t(\alpha)$ to mean the source and target respectively, of the edge $\alpha$)

Question
Can any linear representation of a finite group, $(G,\rho,V)$ be seen as a representation of a quiver $Q$? I think we can take $Q_0=\{1,2\}$ and $Q_1=UG$ where $UG$ is the underlying set of the finite group $G$.
Then, the representation $M$ of $Q$ has $M_1=M_2=V$, and $\varphi_g=\rho(g)$ for each $g\in UG$.
It seems that $gh\in G$ tells us that $\rho(gh)=\rho(g)\rho(h)$ by construction, and my representation of the quiver just looks like a graph with an arrow for each element of $G$ mapping from $V$ to $V$.
In hindsight, perhaps I could take $Q_0=\{1\}$ and $M_1=V$, and take a loop on the one vertex for each element of $UG$.
 A: Strictly speaking, nothing prevents you from doing what you suggest: working over a field $k$, take the quiver $Q$ with vertex set $\{1\}$ and arrow set $\{a_g \ | \ g\in G\}$, subject to the relations $a_ga_h = a_{gh}$ (for all $g,h\in G$) and $a_{1_G} = 1$.  Then the path algebra of $Q$ modulo these relations is isomorphic to the group algebra $kG$.  This implies that the category of representations of the group $G$ is equivalent to the category of representations of $Q$ modulo the above relations.
So why don't people do this?
In one sentence: because quivers are good for studying basic algebras, and $kG$ is not basic in general.
The general idea is the following: If $A$ is an algebra of finite dimension over an algebraically closed field $k$ (this can be more general), then


*

*$A$ is Morita-equivalent to a basic algebra $A_b$, unique up to isomorphism;

*there exists a unique quiver $Q$ and a (non-unique) ideal $I$ of the path algebra $kQ$ such that $A_b$ is isomorphic to $kQ/I$ (to have uniquness of $Q$, we must impose that if $R$ is the ideal generated by the arrows of $Q$, then $R^m\subseteq I \subseteq R^2$ for some $m\geq 2$);

*the representations theory of $Q$  modulo $I$ is well-developed, provided that $I$ satisfies the above property.


For example, if the characteristic of $k$ does not divide the order of $G$, then $kG$ is semisimple, and it is Morita-equivalent (but not isomorphic in general!) to the path algebra of a quiver $Q$ consisting of $m$ vertices and no arrows, where $m$ is the number of irreducible characters of $G$.
Now, the above strategy is sometimes applied to group algebras: allow me to link this answer of mine, which lists a few references.
