Image is G-Module

Theorem
Let $(G,\cdot)$ be a group and let $f:(V,\phi)\to (V^\prime,\mu)$ be an homomorphism of $G$-modules

Then $\operatorname{Im}(f)$ is a $G$-submodule of $V^\prime$.

Proof
From G-Submodule Test we need to proof that $\mu(G,\operatorname{Im}(f))\subseteq \operatorname{Im}(f)$.

In other words: if $g\in G$ and $w\in\operatorname{Im}(f)$, then $\mu(g,w)\in \operatorname{Im}(f)$.

Assume that $g\in G$ and $w\in\operatorname{Im}(f)$, then exists $v\in V$ such that $f(v)=w$.

By definition of homomorphism $\mu(g,w)=\mu(g,f(v))=f(\phi(g,v))\in\operatorname{Im}(f)$.

Thus $\mu(g,w)\in \operatorname{Im}(f)$ and $\operatorname{Im}(f)$ is a $G$-submodule of $V^\prime$