If $\Phi$ is irreducible then $\Delta$ is irreducible.

Question

Let $\Phi$ a root system with basis $\Delta$. Show that if $\Phi$ is irreducible then $\Delta$ is irreducible.

Comments: Suppose that $\Delta = \Delta_1 \cup \Delta_2$ is a partition of $\Delta$ into two non-empty mutually orthogonal subsets. I already showed a result that every root is conjugate to a simple root by an element of the Weyl group $W$ which is generated by the simple reflections. Let $\Phi_1$ the set of roots which are conjugate to an element of $\Delta_1$ and $\Phi_2$ the set those roots which are conjugate to an element of $\Delta_2$. We have $\Phi = \Phi_1 \cup \Phi_2$.

I need to show that $(\Phi_1,\Phi_2) = 0$ to conclude that $\Phi$ is irreducible and get a contradiction.

Answer 1

Suppose $\beta\in\Phi_1$, and $\beta=w(\alpha)$ for $\alpha\in\Delta_1$, $w\in W$. Now $w$ is a product of simple reflections, and the action of a simple reflection $s_i$ corresponding to a simple root $\alpha_i$ in $\Delta_1$ adds a multiple of $\alpha_i$ to $\alpha$, (or whatever root it is after applying some other previous simple reflections in a decomposition of $w$), but the simple reflections for simple roots in $\Delta_2$ leave it unchanged, since $\Delta_1$ and $\Delta_2$ are orthogonal. So $\beta$ is in the span of $\Delta_1$. Likewise, if $\gamma\in\Phi_2$, then $\gamma$ is in the span of $\Delta_2$. Thus $\Phi_1$ and $\Phi_2$ are orthogonal since $\Delta_1$ and $\Delta_2$ are orthogonal.

Answer 2

The first thing you can do is observe that $W\cong W_1\times W_2$ where $W_i=\langle s_\alpha\mid \alpha\in \Delta_i\rangle$ ($i=1,2$). In this case, $W_1$ is the stabilizer of $\Phi_2$, $W_2$ is the stabilizer of $\Phi_1$, and we have $\Phi_1=W_1\Delta_1$ and $\Phi_2=W_2\Delta_2$. It now follows that if $\beta\in\Phi_i$ ($i=1,2$), then $$\beta=\sum_{\alpha\in \Delta_i}c_\alpha \alpha$$ for some $c_\alpha\in \mathbb{R}$. It is now straightforward to conclude that $\Phi_1\perp\Phi_2$.