I'm reading the proof of the fact that if $G$ is a commutative group scheme, its Lie algebra is abelian in Mumford's book abelian varieties chapter 11. It seems to me that an easier proof should work: let $D_1, D_2$ be two left invariant vector fields corresponding to $t_1, t_2: \operatorname{Spec} \Lambda \to G$, where $\Lambda = k[\epsilon]/(\epsilon^2)$ and $t_i$ are tangent vectors of $G$ at $e$. Let $\tilde{D}_i$ be the automorphisms $X \times \operatorname{Spec}\Lambda \to X \times \operatorname{Spec}\Lambda$ induced by $D_i$, then $\tilde{D}_i$ is right-translation by some $t_i \in G(\Lambda)$. But since $G(\Lambda)$ is commutative, $[t_1,t_2]=0$, therefore $[D_1,D_2]=0$.
Is there anything wrong with the above proof? I don't see why you have to base change everything to $\operatorname{Spec}\Lambda'$, where $\Lambda' = k[\epsilon,\epsilon']/(\epsilon^2,\epsilon'^2)$ as done in Mumford.