# Is the longest Weyl transformation of a product group the pair of the longests?

More precisely, if $$G_1,G_2$$ are two Lie groups and $$T_1,T_2$$ are maximal tori respectively, then $$W(G_1\times G_2)\cong W(G_1)\times W(G_2)$$.

The pair of longest elements in $$W(G_1)\times W(G_2)$$ is mapped to some element in $$W(G_1\times G_2)$$ and I would expect this is the longest.

Is this a true statement and can someone show a proof?

If $$\Phi_1$$ and $$\Phi_2$$ denote the root systems of $$G_1$$ and $$G_2$$ w.r.t $$T_1$$ and $$T_2$$, then $$\Phi = \Phi_1 \sqcup \Phi_2$$ is the root system of $$G_1 \times G_2$$. Let us write $$W_i = W(G_i)$$ and set $$W = W_1 \times W_2$$.
Now, we only can speak of longest elements after having fixed sets of simple roots $$\Delta_i \subseteq \Phi_i$$ and then $$\Delta = \Delta_1 \sqcup \Delta_2$$ is a set of simple roots of $$\Phi$$.
So if $$w_i \in W_i$$ denotes the longest element of $$W_i$$ w.r.t. $$\Delta_i$$ we find that $$w_1$$ maps every positive root in $$\Phi_1$$ to a negative one and fixes every root in $$\Phi_2$$ and similarly for the longest element $$w_2$$ of $$W_2$$ w.r.t. $$\Delta_2$$. Thus $$w = w_1 w_2 = w_2 w_1$$ maps every positive root in $$\Phi$$ to a negative one and thus is the longest element.