Image is G-Module

Theorem
Let $\left({G, \cdot}\right)$ be a group and let $f: \left({V, \phi}\right) \to \left({V', \mu}\right)$ be a homomorphism of $G$-modules.

Then $\operatorname{Im} \left({f}\right)$ is a $G$-submodule of $V'$.

Proof
From G-Submodule Test it suffices to prove that $\mu \left({G, \operatorname{Im} \left({f}\right) }\right) \subseteq \operatorname{Im} \left({f}\right)$.

In other words: for any $g \in G$ and $w \in \operatorname{Im} \left({f}\right)$, it is to be shown that $\mu \left({g, w}\right) \in \operatorname{Im} \left({f}\right)$.

Assume that $g \in G$ and $w\in\operatorname{Im} \left({f}\right)$.

Then there exists a $v \in V$ such that $f \left({v}\right) = w$.

By definition of homomorphism, have:


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

Hence, for all $g \in G$ and $w\in\operatorname{Im} \left({f}\right)$, $\mu \left({g, w}\right) \in \operatorname{Im} \left({f}\right)$.

By G-Submodule Test, it follows that $\operatorname{Im} \left({f}\right)$ is a $G$-submodule of $V'$.