If $w'(\beta)<0$ and $\ell(w)+\ell(w')=\ell(ww')$, then $ww'(\beta)<0$? There's a small step in a computation with root systems that eludes me. Suppose $w,w'$ are elements of the Weyl group (which is a Coxeter group) such that $\ell(w)+\ell(w')=\ell(ww')$. Suppose you have positive root $\alpha$, such that $w'^{-1}(\alpha)<0$. Putting $\beta=-w'^{-1}(\alpha)>0$, one has $w'(\beta)<0$, and so
$$
-w(a)=ww'(\beta)<0
$$
since $\ell(w)+\ell(w')=\ell(ww')$. The point is to conclude either $w'^{-1}(\alpha)>0$ or $w(\alpha)>0$.
The one thing I don't understand, why does the additivity of lengths in this case imply that the action of $w$ on $w'(\beta)$ sends it to a negative root?

It is a combination of these two properties:
1) $\ell(uv) \leq \ell(u) + \ell(v)$ for any $u,v \in W$;
2) $\ell(us_\gamma) < \ell(u) \iff u(\gamma) < 0$ for any $u \in W$ and any positive root $\gamma$.
Thus,
\begin{align}
\ell(ww's_\beta) \leq & \; \ell(w) + \ell(ws_\beta) \\\\
< & \; \ell(w) + \ell(w') \\\\
= & \; \ell(ww')
\end{align}
The first inequality is property 1, the strict inequality is property 2 via $w'(\beta) < 0$, and the equality is due to the additivity hypothesis $\ell(ww') = \ell(w) + \ell(w')$. 
Since $\ell(ww's_\beta) < \ell(ww')$, we conclude from property 2 that $ww'(\beta) < 0$, which is equivalent to saying that the action of $w$ on $w'(\beta)$ sends it to a negative root.
