The root sl_2 triple

Choose $0\mathrm{\ne }e\in {\mathfrak{g}}_{\alpha }$, which is possible because $\alpha$ is a root. By [L1] the linear functional $y\mapsto B(e,y)$ on the nonzero space ${\mathfrak{g}}_{-\alpha }$ is not identically zero, hence surjective onto $\mathbb{C}$; choose $f\in {\mathfrak{g}}_{-\alpha }$ with $B(e,f)=2\mathrm{/}\alpha (H_{\alpha })$, a nonzero number by [L3].

For every $H\in \mathfrak{h}$, invariance and the root-space identity give $B([e,f],H)=B(e,[f,H])=\alpha (H)B(e,f)=B(B(e,f)H_{\alpha },H)$. Both $[e,f]$ and $H_{\alpha }$ lie in $\mathfrak{h}$ by [L2], so nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$ yields $[e,f]=B(e,f)H_{\alpha }=\frac{2}{\alpha (H_{\alpha })}{H}_{\alpha }=h_{\alpha }$. By [L4] and [L3], $[h_{\alpha },e]=\alpha (h_{\alpha })e=2e$ and $[h_{\alpha },f]=-2f$. Thus all three bracket relations of The special linear Lie algebra sl_2 hold for $(e,f,h_{\alpha })$.

Since $0\mathrm{\ne }e\in {\mathfrak{g}}_{\alpha }$ and $0\mathrm{\ne }f\in {\mathfrak{g}}_{-\alpha }$ lie in distinct root spaces while $h_{\alpha }\in \mathfrak{h}$, the three elements are linearly independent, so their span is three-dimensional and by step 2.1 is closed under the bracket with the relations of ${\mathfrak{s}\mathfrak{l}}_2$; by The special linear Lie algebra sl_2 it is a copy of ${\mathfrak{s}\mathfrak{l}}_2$. Setting $e_{\alpha }=e$ and $f_{\alpha }=f$ proves the statement.