Proposition 14

Assume that $R$ is irreducible and reduced, and that $(\alpha|\alpha)$ takes the values $\lambda$ and $2\lambda$ for $\alpha \in R$. Let $A$ be the set of roots $\alpha$ such that $(\alpha|\alpha) = \lambda$. Assume that any two non-proportional elements of $A$ are orthogonal. Then $R_1 = R \cup 2A$ is an irreducible non-reduced root system and $R$ is the set of indivisible roots of $R_1$.

It is clear that $R_1$ satisfies (RS$_1$) and (RS$_{III}$). We show that, if $\alpha, \beta \in R_1$, then $\langle \alpha^-, \beta \rangle \in \mathbf{Z}$. This is clear if $\alpha \in R$. Since $(2\alpha)^- = \frac{1}{2} \alpha^-$ for $\alpha \in A$, it is also immediate if $\alpha, \beta \in 2A$. Finally, assume that $\beta \in R$ and that $\alpha = 2\gamma$ with $\gamma \in A$.

1) If $\gamma = \pm \beta$, then $\langle \alpha^-, \beta \rangle = \pm \frac{1}{2} \langle \gamma^-, \gamma \rangle = \pm 1$. 2) If $\gamma$ is not proportional to $\beta$ and if $\beta \in A$, the assumption on $A$ implies that $\langle \gamma^-, \beta \rangle = 0$, so $\langle \alpha^-, \beta \rangle = 0$. 3) If $\beta \in R - A$, then $(\beta|\beta) = 2\lambda = 2(\gamma|\gamma)$, so $\langle \beta, \gamma^- \rangle$ is equal to 0, 2 or -2 by the list in no. 3. Thus $\langle \beta, \alpha^- \rangle = \frac{1}{2} \langle \beta, \gamma^- \rangle \in \mathbf{Z}$.

Thus $R_1$ is a root system in $V$, and the other assertions are clear.