G-Submodule Test

Theorem

Let $\struct {V, \phi}$ be a $G$-module over a field $k$.

Let $W$ be a vector subspace of $V$.

Let $\phi_W: G \times W \to V$ denote the restriction of $\phi$ to $G \times W$.

Then:

$\struct {W, \phi_W}$ is a $G$-submodule of $V$

if and only if:

$\map \phi {G, W} \subseteq W$

Proof

Necessary Condition

Let $W$ be a $G$-submodule of $V$.

Hence by definition $\phi_W: G \times W \to W$ is a linear action on $W$.

Also by definition:

$\map {\phi_W} {G, W} = \map \phi {G, W} \subseteq W$

$\Box$

Sufficient Condition

Let:

$\map \phi {G, W} = \map {\phi_W} {G, W} \subseteq W$

We have that $\phi_W: G \times W \to W$ is a well-defined mapping.

We need to check if $\phi_W$ is a linear action on $W$:

Assume $a, b \in W$ and $g \in G$.

In particular:

$a, b \in V$

and so:

\(\ds \map {\phi_W} {g, a + b}\)

\(=\)

\(\ds \map \phi {g, a + b}\)

Definition of $\phi_W$

\(\ds \)

\(=\)

\(\ds \map \phi {g, a} + \map \phi {g, b}\)

$\phi$ is a linear action on $V$

\(\ds \)

\(=\)

\(\ds \map {\phi_W} {g, a} + \map {\phi_W} {g, b}\)

Definition of $\phi_W$

Further, let $\lambda \in k$ and $g \in G$.

Then:

\(\ds \map {\phi_W} {g, \lambda b}\)

\(=\)

\(\ds \map \phi {g, \lambda b}\)

Definition of $\phi_W$

\(\ds \)

\(=\)

\(\ds \lambda \map \phi {g, b}\)

$\phi$ is a linear action on $V$

\(\ds \)

\(=\)

\(\ds \lambda \map {\phi_W} {g, b}\)

Definition of $\phi_W$

Thus $W$ is a $G$-submodule of $V$.

$\blacksquare$