 Solution

Use the nondegenerate restriction of the Killing form to $\mathfrak{t}$ to define $t_{\alpha }\in \mathfrak{t}$ byKilling-form invariance and the root-space decomposition show that ${\mathfrak{g}}_{\alpha }$ pairs nondegenerately with ${\mathfrak{g}}_{-\alpha }$ and orthogonally with every other root space. Choose nonzero $e\in {\mathfrak{g}}_{\alpha }$ and $f\in {\mathfrak{g}}_{-\alpha }$ with $\kappa (e,f)\ne 0$. For $t\in \mathfrak{t}$,so

We need $\alpha (t_{\alpha })\ne 0$. If it were zero, the span of $e,f,t_{\alpha }$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By Lie theorem, its adjoint action on $\mathfrak{g}$ can be upper triangularized, so $\mathrm{ad}[e,f]$ is nilpotent. But $[e,f]\in \mathfrak{t}$, and elements of the Cartan subalgebra act semisimply; hence $\mathrm{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne 0$.

SetRescale $f$ so that $[e_{\alpha },f_{\alpha }]=h_{\alpha }$. Since $e_{\alpha }$ and $f_{\alpha }$ lie in the $\alpha$ and $-\alpha$ root spaces,Thus ${\mathfrak{m}}_{\alpha }=⟨e_{\alpha },h_{\alpha },f_{\alpha }⟩$ is the sl2 subalgebra associated with a root.