Let $K$ be a field, $G$ a group and $KG$ the corresponding group ring. Let $M$ be a $KG$-module, therefore it can also be viewed as a $K$-vector space. Suppose that $M$ has finite $K$-dimension, $\dim_KM = n$. Then, $\exists U \leq M$ a $KG$-submodule of $M$ such that $U$ is simple.
Is this result true?
On
I arrived at the following proof, just wanted to make sure it is fine.
Proof.
If $M$ is not simple it means that $\exists T_1 < M$ a non-zero proper submodule of $M$. Since $T$ is a $KG$-submodule of $M$ it is also a $K$-subspace of $M$ viewed as a $K$-vector space. We have that $\dim_k T_1 = n_1 \leq n$. If $n_1 = n$ it would mean that $T_1 = M$ when viewing them as vector spaces. Thus, $n_1 < n$.
If $T_1$ is not simple we may apply the same argument inductively in at most $n$ steps, until arriving at a $T_m$ such that $T_m$ is simple or has $\dim_KT_m = 1$. In case that $\dim_kT_m =1$, suppose that $T \leq T_m$ is a submodule, then it is also a $K$-subspace of $T_m$. However, since $T_m$ is of dimension $1$, its only subspaces are $0$ and itself, thus $T_m$ is simple.
We have found $T_m$ as a simple submodule of $M$.
You can simply say that because $M$ is finite dimensional, it is also apparently Artinian, and so the collection of all nonzero submodules must contain a minimal element.