TITLE: Inversion set determine an element in a symmetric group.
QUESTION [5 upvotes]: Let $S_n$ be the symmetric group over $\{1,2,\ldots,n\}$. The inversion set of a element $w$ in $S_n$ is defined by $Inv(w)=\{(i,j): i<j, w^{-1}(i)>w^{-1}(j)\}$. I think that if $w, w' \in S_n$ and $Inv(w)=Inv(w')$, then $w = w'$. Is this true? Thank you very much.

REPLY [3 votes]: I use the definition $Inv(w) = \{ (i,j) \mid i < j \text{ and } w(i) > w(j) \}$, see Derek Holt's comment.
Giving a permutation $w$ is the same as giving a total order $<_w$, i.e. $w(1) <_w w(2) <_w ... <_w w(n)$, on the set $\{1,...,n\}$.
On the other hand the relation $<_w$ is completely determined by the set $Inv(w)$: For $i \neq j$ we have
\begin{align}
i <_w j
\iff& \left( i > j \text{ and } w(i) < w(j) \right) \text{ or } \\
    & \left( i < j \text{ and } w(i) < w(j) \right) \\
\iff& (j,i) \in Inv(w) \text{ or } \\
    & \left( i < j \text{ and } (i,j) \notin Inv(w) \right)
\end{align}
In summary we have bijections between:
$$ S_n \overset{1:1}\leftrightarrow \text{total orders on [n]} \overset{1:1}\leftrightarrow \text{inversion sets}$$