= Solution
If the three <simple roots> are distinct and every pair of root spaces had nonzero <Lie bracket>, part (iv) would put all three connections in the <Dynkin diagram>. Their underlying <Coxeter graph> would contain a triangle, contradicting part (ii). Hence \b[at least one pair satisfies $\boxed{[\mathfrak g_{\alpha_i},\mathfrak g_{\alpha_j}]=0}$.] If two of the listed roots coincide, their common <root space> is one-dimensional, so its bracket with itself is already zero. Thus the conclusion also covers that case.
Back to article page
