Disclaimer: This question was first posted on math.se without any answer.

This is something that naturally occurs in my research, but I am no expert on this - it feels like a natural question so I am hoping for a reference or a short proof.

Consider braids on $n$ strands with the group generated by $s_1,\dotsc, s_{n-1}$, *left-twists*, and their inverses, *right twists*.

Remember the classical braid relations, $s_i s_j = s_j s_i$ if $|i-j| \geq 2$, and $s_i s_{i+1}s_i=s_{i+1} s_{i}s_{i+1}$.

Let $R(w)$ be all reduced words for an element $w$ in the braid group, and let $RT(w)$ be defined as the maximal positive exponent that appears, among all representatives of $w$ in $R(w)$. That is, $RT(w)$ somehow captures the maximal number of consecutive right twists among all representatives of $w$, without cheating by introducing more right twists followed by same number of left twists (remember, we only consider reduced words).

**Q1:** I would like to prove that $RT(uw) \leq RT(u)+RT(w)$.

**Q2:** The special case that I really care about arises from permutations: let $u$ and $w$ be reduced words of *permutations* in $S_n$, meaning no $s_i^2$ appears.

Re-interpret these two words as elements in the braid group (by doing right-twists) and consider the braid $b = uw^{-1}$,
a bunch of right-twists followed by left-twists.

Is it true that for any *reduced* representative of $b$ in the braid group, there is no exponent greater than $1$, i.e.,
$s_i^2$ is not a subword?

In English - can left-twists increase the number of right-twists?

*Note:* Of course $uw^{-1}$ might not be reduced, one can use braid relations to cancel some left-twists with right twists.