Corollary

Let $(\mathfrak{g},\mathfrak{h})$ be a split semi-simple Lie algebra. Then $(\mathfrak{g},\mathfrak{h})$ possesses a Chevalley system (§2, no. 4, Def. 3).

Let R be the root system of $(\mathfrak{g},\mathfrak{h})$. For all $\alpha \in R$, let $X_{\alpha}$ be a non-zero element of $\mathfrak{g}^{\alpha}$. Assume that the $X_{\alpha}$ are chosen so that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ for all $\alpha \in R ($§2, no. 4, Lemma 2). Let B be a basis of R and $\theta$ the automorphism of $\mathfrak{g}$ such that $\theta (X_{\alpha}) =X_{-\alpha}$ for all $\alpha \in B\cup (-B)$. We have $\theta |\mathfrak{h}=-$Id$_{\mathfrak{h}}$. Hence, for all $\alpha \in R$ there exists $t_{\alpha}\in k^*$ such that $\theta X_{\alpha}=t_{\alpha}X_{-\alpha}$. We have

so $t_{\alpha}t_{-\alpha}= 1$ for all $\alpha \in R$. Introduce the $N_{\alpha \beta}$ as in §2, no. 4. If $\alpha , \beta , \alpha +\beta \in R$,

so, in view of §2, no. 4, Lemma 4,

where $q$ is an integer. It follows that if $t_{\alpha}$ and $t_{\beta}$ are squares in $k^*$, so is $t_{\alpha+\beta}$. Since $t_{\alpha}= 1$ for all $\alpha \in B$, Prop. 19 of Chap. VI, §1, no. 6, proves that $t_{\alpha}$ is a square for all $\alpha \in R$. Choose, for all $\alpha \in R$, a $u_{\alpha}\in k$ such that $u^2_{\alpha}=t_{\alpha}$. This choice can be made so that $u_{\alpha}u_{-\alpha}= 1$ for all $\alpha \in R$. Put $X'_{\alpha}=u^{-1}_{\alpha}X_{\alpha}$. Then, for all $\alpha \in R$,

and $\theta (X'_{\alpha}) =\theta (u^{-1}_{\alpha}X_{\alpha}) =u^{-1}_{\alpha}t_{\alpha}X_{-\alpha}=u_{\alpha}X_{-\alpha}=u_{\alpha}u_{-\alpha}X'_{-\alpha}=X'_{-\alpha}$, so that $(X'_{\alpha})_{\alpha\in R}$ is a Chevalley system of $(\mathfrak{g},\mathfrak{h})$.