# Commutative Linear Transformation is G-Module Homomorphism

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## 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 Correspondence between Linear Group Actions and Linear Representations:

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.

$\blacksquare$