I have been trying to show that the radical of $\mathfrak{gl}(2,\mathbb{C})$ is its center, i.e. scalar matrices, however all the proofs I have encountered (e.g. Radical of $\mathfrak{gl}_n$) have used the fact that $\mathfrak{sl}(2,\mathbb{C})$ is semisimple. Instead, I am aiming to find the radical of $\mathfrak{gl}(2,\mathbb{C})$ first, and then use the fact that $\mathfrak{gl}(2,\mathbb{C}) \cong \mathfrak{sl}(2,\mathbb{C}) \bigoplus \mathbb{C}$ along with the proposition:
- If $L$ is a Lie algebra, then $L/Rad(L)$ is semisimple,
to conclude that $\mathfrak{sl}(2,\mathbb{C})$ is indeed semisimple.
Clearly we have that $Z(\mathfrak{gl}(2,\mathbb{C})) \subseteq Rad(\mathfrak{gl}(2,\mathbb{C}))$, however I am not sure how to even begin showing the opposite inclusion.
Any help would be much appreciated.
It is quite simple to show that $\mathfrak{sl}_2(\mathbb{C})$ is simple (not just semisimple) by performing an explicit calculation so I'm not sure what you gain by going the way you suggest.
In any case, let $\mathfrak{g} = \mathfrak{gl}_2(\mathbb{C})$. If $\mathrm{Rad}(\mathfrak{g})$ is three dimensional, then the quotient $\mathfrak{g} / \mathrm{Rad}(\mathfrak{g})$ is one dimensional, hence abelian and solvable. If $\mathrm{Rad}(\mathfrak{g})$ is two dimensional, then the quotient $\mathfrak{g} / \mathrm{Rad}(\mathfrak{g})$ is two dimensional, hence solvable (there are only two two-dimensional Lie algebras up to isomorphism - one abelian, the other non-abelian but solvable). Hence, in both cases, we would have that $\mathfrak{g}$ is solvable which is clearly false.