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

Theorem
Let $\left({G, \cdot}\right)$ be a finite group.

Let $\left({V, \phi}\right)$ be a $G$-module.

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

Proof
Assume that $V$ is an irreducible $G$-module, but it has a proper $G$-submodule.

By the definition irreducible then its associated representation (let's call it $\tilde{\phi}=\rho:G\to \operatorname{GL}(V)$) is irreducible.

In Equivalence of Representation Definitions it is defined $\rho(g)(v)=\phi(g,v)$ where $g \in G$ and $v\in V$.

Since $V$ has a proper $G$-submodule, there exists $W$ a proper  vector subspace which $\phi\left({G,W}\right)\subseteq W$ and so $\rho\left({G}\right)W\subseteq W$.

Hence $W$ is invariant by every linear operators in $\left\{{\rho(g): g \in G}\right\}$.

By definition, $\rho$ cannot be irreducible.

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

Assume now that $V$ has no proper $G$-submodules, but it is reducible $G$-module.

By the definition of reducible $G$-module follows that its associated representation (let's call it $\tilde{\phi}=\rho:G\to \operatorname{GL}(V)$) is reducible.

From the definition of reducible representation follows that there exists a vector space $W$ of $V$ which is invariant under all the linear operators in $\{\rho(g):g\in G\}$.

Then $\phi(G,W)=\rho(G)W\subseteq W$; which is the definition of $G$-submodule of $V$.

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

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