I am currently failing to understand a statement in a proof that is (maybe) considered too trivial to explain.
Let $K$ be a field of characteristic zero, and $\mathfrak{g}$ be a Lie-subalgebra of $M_2(K)$ satisfying the following properties:
1) $K^2$ is an irreducible $\mathfrak{g}$-module
2) The centralizer of $\mathfrak{g}$ in $M_2(K)$ is $K$, i.e. the only matrices commuting with all matrices in $\mathfrak{g}$ are of the form $\lambda E_2$, $E_2$ being the identity matrix.
Then, $\mathfrak{g}$ is either $M_2(K)$ or $\mathfrak{sl}_2$.
I have no idea how to prove this, or how to find literature that covers this and would appreciate any help :)
By $(2)$, the center of $\mathfrak{g}$ is at most $1$-dimensional. Since irreducible modules of solvable Lie algebras are $1$-dimensional by Lie's Theorem, $(1)$ says that $\mathfrak{g}$ is a non-solvable Lie algebra of dimension $n\le 4$. Together this leaves only $\mathfrak{sl}_2(K)$ and $\mathfrak{gl}_2(K)$.