Exercise 3 in Chapter IV, Section 1 on Bourbaki, *Lie Groups and Lie Algebras* This is exercise 3 in Bourbaki, Lie Groups and Lie Algebras.  The application I'm interested in is the fact that for any $\theta \subseteq \Delta$ in a reduced root system, there exists a unique left coset representative of each left coset $W/W_{\theta}$ with minimal length, and it has several nice properties which I want to understand better.

 A: Let $X = \{r_i\}_{i \in I}$ and $Y = \{u_j\}_{j \in J}$.
First, note $W_X w' W_Y =  W_X a W_Y$ for any $w' \in W_X a W_Y$ since $W_X$ and $W_Y$ are groups. In each of our arguments, we will make use of the following lemma.
Lemma:   There is a unique element $w \in W_X a W_Y$ such that $\ell(r w) = \ell(w u) = \ell(w) + 1$ for all $r \in X, u \in Y$. 
Proof: Let $v,w$ both satisfying the condition of the lemma. Then $v = x w y$ for some $x \in W_X$ and $y \in W_y$. Let $r_1 \dots r_k$, $s_1 \dots s_l$ and $u_1 \dots u_m$ be reduced expressions of $x$, $w$ and $y$ respectively. Then there is a subword of $r_1 \dots r_k s_1 \dots s_l u_1 \dots u_m$ that is a reduced expression of $v$. The reduced expression of $v$ cannot begin with $r_i$ or end in $u_j$, else $\ell(r_i v),\ell(v u_j) < \ell(v)$, so $v = s_1 \dots s_l = w$.
The length function $\ell$ of $W$ gives a grading of $W_X a W_y$, so there must exist at least one minimal element $w$. For any minimal element $v$, we see $\ell(r v) = \ell(v u) = \ell(v) + 1$ for all $r \in X, u \in Y$. Then by the Lemma, $w$ must be the unique minimal element.
Next we show that for each $w' \in W_X a W_Y$, there exist $x \in W_X$ and $y \in W_Y$ such that $w' = xwy$ with $\ell(w') = \ell(x) + \ell(w) + \ell(y)$. We proceed by induction on $\ell(w') - \ell(w)$. Clearly the result holds when $w' = w$, so assume $w' \neq w$. By the Lemma, we see there exists $r_i \in X$ (or $u_j \in Y$) such that $\ell(r_i w') = \ell(w') - 1$ (or $\ell(w' u_j) = \ell(w') - 1$), else $w' = w$. Then $r_i w' = x w y$ so $w' = r_i x w y$ with $\ell(w') = 1 + \ell(x) + \ell(w) + \ell(y)$. Thus $\ell(r_i x) = 1 + \ell(x)$, completing our proof.
The cases where $X = \varnothing$ and $Y = \varnothing$ follow as corollaries of the above work. The fact that $w$ is $(X,Y)$-reduced if and only if $w$ is $(X,\varnothing)$-reduced and $(\varnothing,Y)$-reduced follows from the Lemma, since in both cases $\ell(r_i w) = \ell(w u_j) = \ell(w) + 1$ for all $i \in I$ and $j \in J$.
