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

Theorem
Let $(G,\cdot)$ be a finite group and let $(V,\phi)$ be a $G$-module.

Then $V$ is an irreducible $G$-module iff $V$ has no 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 $\phi(G,W)\subseteq W$ and so $\rho(G)(W)\subseteq W$, hence $W$ is invariant by every linear operators in $\{\rho(g):g\in G\}$.

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 a reducible $G$-module.