Krdytkyu

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.

Suppose $betainPhi_1$, and $beta=w(alpha)$ for $alphainDelta_1$, $win W$. 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 $gammainPhi_2$, 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.

The first thing you can do is observe that $Wcong W_1times W_2$ where $W_i=langle s_alphamid alphain Delta_irangle$ ($i=1,2$). In this case, $W_1$ is the stabilizer of $Phi_2$, $W_2$ is the stabilizer of $Phi_1$, and we have $Phi_1=W_1Delta_1$ and $Phi_2=W_2Delta_2$. It now follows that if $betainPhi_i$ ($i=1,2$), then
$$beta=sum_{alphain Delta_i}c_alpha alpha$$
for some $c_alphain mathbb{R}$. It is now straightforward to conclude that $Phi_1perpPhi_2$.