Suppose $L$ is a finite dimensional Lie algebra. Let $V$ be an $L$-module (i.e. $V$ is a vector space which $L$ acts upon). We are assuming that $V$ has a finite dimension. My question is the following:
Does $V$ necessarily contain an irreducible $L$-submodule?
For reference, irreducible $L$-module is one which has only two $L$-submodules, namely $0$ and itself. By the way, $0$ subspace does not count as an irreducible submodule (following the convention in Humphreys' Introduction to Lie Algebras and Representation Theory).
I suspect that the answer is 'yes', but perhaps a clever counterexample exist?
The answer is yes, every finite dimensional $L$-module, $V,$ contains an irreducible submodule. The proof is by induction on the dimension of $V.$
The base case is clear since any $1$ dimensional module is automatically irreducible. For the induction step assume $\dim V > 1.$ If $V$ is irreducible we are done so suppose it is not. Then by definition of irreducible $V$ contains a proper submodule $W\subset V.$ By the induction assumption $W$ contains an irreducible submodule hence so does $V.$