Commutative Linear Transformation is G-Module Homomorphism

Theorem
Let $\rho: G \to \operatorname{GL} \left({V}\right)$ be a representation.

Let $f: V \to V$ be a linear mapping.

Let:
 * $\forall g \in G: \rho \left({g}\right) \circ f = f \circ \rho \left({g}\right)$

Then $f: V \to V$ is a $G$-module homomorphism.

Proof
Let:
 * $\forall g \in G: \rho \left({g}\right) \circ f = f \circ \rho \left({g}\right)$

Let $v$ be a vector $v \in V$.

Then:
 * $\rho \left({g}\right) \left({f \left({v}\right)}\right) = f \left({\rho \left({g}\right) \left({v}\right)}\right)$

Using the properties from Existence of Bijection between Linear Group Action and Linear Representation:
 * there exists a $G$-module $\left({V, \phi}\right)$ associated with $\rho$ such that:
 * $\phi \left({g, v}\right) = \rho \left({g}\right) \left({v}\right)$

Applying the last formula:


 * $\rho \left({g}\right) \left({f \left({v}\right)}\right) = \phi \left({g, f \left({v}\right)}\right)$

and:


 * $f \left({\phi \left({g, v}\right)}\right) = f \left({\rho \left({g}\right) \left({v}\right)}\right)$

Thus our assumption is equivalent to:
 * $f \left({\phi \left({g, v}\right)}\right) = \phi \left({g, f \left({v}\right)}\right)$

Hence, by definition of $G$-module homomorphism, $f: V \to V$ is a $G$-module homomorphism.