I can't see why a claim I'm reading is true.
If $\sigma\in S_n$, let $R(\sigma)=\{(i,j):i<j,\ \sigma(i)>\sigma(j)\}$, i.e., $R(\sigma)$ is the set of inversions of $\sigma$.
The set $C_n=\{1,(n-1,n),(n-2,n-1,n),\dots,(1,2,\dots,n)\}$ is a set of coset representatives of $S_n/S_{n-1}$. The claim is that if $\sigma\in S_{n-1}$ and $g\in C_n$, then $$ R(g\sigma)=R(\sigma)\sqcup R(g) $$ in order to show $\ell(g\sigma)=\ell(g)+\ell(\sigma)$. Here $\ell(\sigma)$ is the length of $\sigma$, which is the minimal number of transpositions needed to write $\sigma$, which is also known to be equal to the number of inversions of $\sigma$.
I don't see why $R(g\sigma)=R(\sigma)\sqcup R(g)$. For instance, $g$ has form $(n-j,n-j+1,\dots,n)$, so the inversions are precisely the pairs $(n-k,n)$ for $j\leq k\leq 1$.
So supposedly $(n-k,n)\in R(g\sigma)$ as well? First, $g\sigma(n)=g(n)=n-j$. But for $g\sigma(n-k)$, what if $\sigma$ sends $n-k$ to some element less than $n-j$, then $g$ fixes it, so $g\sigma(n-k)<n-j=g\sigma(n)$, so $(n-k,n)\notin R(g\sigma)$.
Maybe the equality of inversions is wrong, but the equation of lengths is still correct?
Source: Quiver Hecke Algebras and 2-Lie Algebras It's at the bottom of page 3, following Proposition 2.2.