Kernel is G-Module

From ProofWiki
Jump to navigation Jump to search

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 $V$.

$\blacksquare$