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
We need to proof 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 $v\in\ker(f)$, then $f(\phi(g,v))=\mu(g,f(v))=\mu(g,0)=0$. Thus $\phi(g,v)\in \ker(f)$ and $\ker(f)$ is a $G$-submodule of $V$