This is found in the book Linear Algebra by Hoffman and Kunze (Ch. 5.7 The Grassman Ring, last example). A bit of context : Here we are talking about Laplace Expansions, which shows that a determinant of an $n$ x $n$ matrix can be acquired from the wedge product of two other determinants of $r$ x $r$ and $s$ x $s$ matrix with $r + s = n$ and $r$ strictly smaller then $n$. But here in order to express it 'explicitly' we need to calculate $\operatorname{sgn}{\sigma}$ where $\sigma$ is picked one from each coset of $G$ defined below:
The fact that $\sigma$ with the special property stated above is unique for each left coset of G is clear to me, but how do we compute them explicitly as stated above? Thank you!
Let's use the fact that the sign of a permutation is given by $sgn(\sigma) = (-1)^N$, where $N$ is the number of inversions of $\sigma$, where an inversion of $\sigma$ is a pair of numbers $(i,j)$ such that $i<j$ and $\sigma(i) > \sigma(j)$.
Let's consider a permutation $\sigma$ of the form which you describe: $\sigma 1 < \dots < \sigma r$, $\sigma (r+1) < \dots < \sigma n$. The inversions must be of the form $(i, j)$ where $i \in \{1, \dots, r \}$ and $j \in \{r+1, \dots, n \}$. Specifically for each $i \in \{1, \dots, r\}$ the number of inversions which $i$ belongs to is $\sigma i - i$. So the total number of inversions is $\sigma 1 + \dots \sigma r - r(r-1)/2$, so the sign of the permutation is given by $(-1)^{\sigma 1 + \dots \sigma r + r(r-1)/2}$.