G-Module is Irreducible iff no Non-Trivial Proper Submodules

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Theorem

Let $\struct {G, \circ}$ be a finite group.

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

Then $V$ is an irreducible $G$-module if and only if $V$ has no non-trivial proper $G$-submodules.

Proof

Necessary Condition

Aiming for a contradiction, suppose that $V$ is an irreducible $G$-module, but it has a non-trivial proper $G$-submodule.

By the definition irreducible, its associated representation is irreducible.

Let this representation be denoted:

$\tilde \phi = \rho: G \to \GL V$

In Correspondence between Linear Group Actions and Linear Representations it is defined as:

$\map {\map \rho g} v = \map \phi {g, v}$

where $g \in G$ and $v \in V$.

Since $V$ has a proper $G$-submodule, there exists $W$ a non-trivial proper vector subspace such that:

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

and so: $\rho \sqbrk G W \subseteq W$

Hence $W$ is invariant by every linear operators in $\set {\map \rho g: g \in G}$.

By definition, $\rho$ cannot be irreducible.

Thus we have reached a contradiction, and $V$ has then no non-trivial proper $G$-submodules.

$\Box$

Sufficient Condition

Aiming for a contradiction, suppose now that $V$ has no proper $G$-submodules, but it is a reducible $G$-module.

By the definition of reducible $G$-module, it follow that its associated representation is reducible.

Let this representation be denoted:

$\tilde \phi = \rho: G \to \GL V$

From the definition of reducible representation, it follows that there exists a vector space $W$ of $V$.

This is invariant under all the linear operators in $\set {\map \rho g: g \in G}$.

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Then:

$\phi \sqbrk {G, W} = \rho \sqbrk G W \subseteq W$

which is the definition of a $G$-submodule of $V$.

By our assumption, $V$ has no non-trivial proper $G$-submodules.

Thus we have reached a contradiction and $V$ must be then an irreducible $G$-module.

$\blacksquare$