Kernel is G-Module
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Theorem
Let $\struct {G, \cdot}$ be a group.
Let $f: \struct {V, \phi} \to \struct {V', \mu}$ be a homomorphism of $G$-modules.
Then its kernel $\map \ker f$ is a $G$-submodule of $V$.
Proof
From G-Submodule Test it suffices to prove that $\phi \sqbrk {\struct {G, \map \ker f} } \subseteq \map \ker f$.
That is, it is to be shown that, if $g \in G$ and $v \in \map \ker f$, then $\map \phi {g, v} \in \map \ker f$.
Assume that $g \in G$ and $v \in \map \ker f$.
\(\ds \map f {\map \phi {g, v} }\) | \(=\) | \(\ds \map \mu {g, \map f v}\) | $f$ is a $G$-module homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \mu {g, 0}\) | $v \in \map \ker f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | $\mu$ is a linear action |
Thus $\map \phi {g, v} \in \map \ker f$.
Hence $\map \ker f$ is a $G$-submodule of $V$.
$\blacksquare$