 Solution

The root-space decomposition gives $[L_r,L_s]\subseteq L_{r+s}$ when $r+s\ne 0$. The two roots are independent, so $r+s\ne 0$, and the missing root sum means that this weight space is zero. Hence $[e_r,e_s]=0$. These adjoint root operators are nilpotent: successive brackets shift root weights by the same root, so finite root strings force eventual vanishing; on the Cartan subalgebra the first bracket lies in its root space and the next is zero. By part (iv), the nilpotent operators $A=t\mathrm{ad}{e}_r$ and $B=u\mathrm{ad}{e}_s$ commute. Every power of $A$ therefore commutes with every power of $B$, so their finite matrix exponentials commute too. Using the inverses proved in part (ii),