What does it mean if a representation "factors through a group"?

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I encountered this in this paper:

Suppose $\rho$ factors through a finite cyclic group $G = \operatorname{Gal}(F/K) \simeq C_n$.

I know a little bit about representation theory, but unfortunately, I have never heard of factoring through a group before, so I do not know what exactly this is supposed to mean. Could you please explain that to me? Thank you!

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Generally speaking, you can see (or even define) a representation $\rho$ of a group $G$ as a group homomorphism $G\to GL(V)$ for some vector space $V$. Then saying that the representation factors through a group $H$ means that this group homomorphism factors through $H$, in the sense that there exist group homomorphisms $\varphi: G\to H$ and $\psi: H \to GL(V)$ such that $\rho=\psi\circ \varphi$.

If you prefer to see a representation as a certain function $G\times V\to V:(g,v) \mapsto g\ast_\rho v$ satisfying certain conditions, then the above tells you that $g\ast_\rho v=\varphi (g)\ast_\psi v$ for all $g,v$.

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The reason for this "factoring" terminology is that homomorphisms between groups behave, to a certain extent, like multiplication: there is a binary operation called "composition"; and this operation satisfies the associative law. For example, we teach our students to factor functions as compositions in order to apply the chain rule: $f(x)=\cos(x^2)$ factors as $f=g \circ h$ where $h(x)=x^2$ and $g(x)=\cos(x)$.

However, one way this differs from multiplication (besides the failure of commutativity) is that the domains and ranges need not all be the same. The point of "factoring through $G$" is that one can factor the representation so that the group in the middle is $G$. Denoting your given group representation as $\rho : K \to \Gamma$ (where $\Gamma$ might be the general linear group of some vector space, or a Lie group, or any other kind of group), this means that there are two group homomorphisms $$K \xrightarrow{\sigma} G \xrightarrow{\tau} \Gamma $$ such that $\rho = \tau \circ \sigma$.