Cartan's semisimplicity criterion

Suppose $\mathfrak{g}$ is semisimple and let $\mathfrak{a}={\mathfrak{g}}^{\perp }$ be the radical of $K$. It is an ideal by [L1]. For $x,y\in \mathfrak{a}$, the operator ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and induces zero on $\mathfrak{g}\mathrm{/}\mathfrak{a}$; its trace on $\mathfrak{g}$ therefore equals the trace of its restriction to $\mathfrak{a}$. Thus the intrinsic Killing form of $\mathfrak{a}$ is the restriction of $K$, hence zero.

Conversely suppose $K$ is nondegenerate and let $\mathfrak{a}$ be an abelian ideal. For $a\in \mathfrak{a}$ and $x\in \mathfrak{g}$, put $T={\mathrm{ad}}_a{\mathrm{ad}}_x$. Its image lies in $\mathfrak{a}$, while $T$ vanishes on $\mathfrak{a}$ because $[x,\mathfrak{a}]\subseteq \mathfrak{a}$ and $\mathfrak{a}$ is abelian. Hence $T^2=0$, so $K(a,x)=\mathrm{tr}(T)=0$. Nondegeneracy forces $a=0$; thus $\mathfrak{g}$ has no nonzero abelian ideal.

By [L2], step 1.1 makes $\mathfrak{a}$ solvable. It is a solvable ideal of the semisimple algebra $\mathfrak{g}$, so [L3] gives $\mathfrak{a}=0$. Therefore $K$ is nondegenerate.

If $\mathfrak{g}$ had a nonzero solvable ideal $\mathfrak{r}$, let ${\mathfrak{r}}^{(m)}$ be the last nonzero term of its derived series: it is nonzero and abelian, and it is an ideal of $\mathfrak{g}$ because derived terms of an ideal are ideals of the ambient algebra by Jacobi. This contradicts step 1.2. Thus the radical is zero and [L3] makes $\mathfrak{g}$ semisimple. For $\mathfrak{g}=0$, the unique bilinear form has zero radical and is nondegenerate in the standard vacuous sense, so both directions still hold.