Let $S_n$ be the symmetric group on the set $\{1,\dots, n\}$ and $\sigma\in S_n$. An inversion of $\sigma$ is a pair $(i,j)$ such that $1\leq i<j\leq n$ and $\sigma(i)>\sigma(j)$. Let $S_n$ be generated as a Coxeter group by the reflections $s_i=(i,i+1)$, $i=1,\dots, n-1$. I want to prove that the number of inversions of $\sigma$, $I_{\sigma}$, equals its Coxeter lenght, $l$.
I've thought of a proof by induction on the length $l$. If $l=0$, $\sigma=()$ son the number of inversions is $0=l$. Assume that for $l=k-1$ the claim holds, and let $\sigma=s_{i_1}\cdots s_{i_k}=s_{i_1}\tau$. We know that the number of inversions of $\tau$ equals $k-1$, and that $I_{\sigma}-I_{\tau}$ is an odd number because $sign(\sigma)=-sign(\tau)$.
How can I prove that the difference is precisely $1$?
Edit: I can show that $I_{\sigma}=I_{\tau}\pm 1$ because any transposition changes the number of inversiones by $\pm 1$. But now I'm stuck in how to prove that it must be $+1$.
There is a proof of this in Combinatorics of Coxeter Groups by Bjorner and Brenti (Proposition 1.5.2).
Here is it roughly. Let $L(\sigma)$ be the length of a permutation in the Coxeter group $A_{n-1}$. Let $$Inv(\sigma) = \#\{(i,j) : i < j \text{ and } \sigma_i > \sigma_j \}$$ be the number of inversions. Let $s_i$ be the transposition which swaps $i$ and $i+1$.
Let $$\sigma = s_{i_1}s_{i_2}\ldots s_{i_{L(\sigma)}}$$ be a minimal length decomposition of $\sigma$. As you noted for all permutations $\pi$, $Inv(\pi s_i) = Inv(\pi) \pm 1$ based on if $\pi < \pi_{i+1}$ or not. So $Inv(s_{i_1}s_{i_2}\ldots s_{i_{L(\sigma)}}) \leq L(\sigma)$.
If $Inv(\sigma) = 0$ then $\sigma = e$. We induct on the number of inversions. Assume for all $\sigma$ with $Inv(\sigma) \leq k$ that $L(\sigma) \leq Inv(\sigma). $If $Inv(\sigma) = k+1$, since $\sigma$ is not the identity there exists $i$ such that $\sigma_i \not < \sigma_{i+1}$ and hence $Inv(\sigma s_i) = Inv(\sigma) - 1$