Kernel 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 the kernel of $f$, $\ker \left({f}\right)$, is a $G$-submodule of $V$.

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

That is, it is to be shown that, if $g \in G$ and $v \in \ker \left({f}\right)$, then $\phi \left({g, v}\right) \in \ker \left({f}\right)$.

Assume that $g \in G$ and $v \in \ker \left({f}\right)$.

Thus $\phi \left({g, v}\right) \in \ker \left({f}\right)$.

Hence $\ker \left({f}\right)$ is a $G$-submodule of $V$.