Kernel 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 $\ker(f)$ is a $G$-submodule of $V$.

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

In other words: if $g\in G$ and $v\in\ker(f)$, then $\phi(g,v)\in \ker(f)$.

Assume that $g\in G$ and $v\in\ker(f)$.

Thus $\phi(g,v)\in \ker(f)$ and $\ker(f)$ is a $G$-submodule of $V$