Must two permutations in $S_n$ that only differ at two adjacent positions have different signs (i.e., one is even and one is odd)?

Question

The Problem: Must two permutations in $S_n$ that only differ at two adjacent positions have different signs (i.e., one is even and one is odd)?

For example, suppose $\sigma, \tau\in S_n$ such that $\sigma(x)=\tau(x) \forall x\in[n]-\{i, i+1\}$, $\sigma(i)=\tau(i+1)=a$ and $\sigma(i+1)=\tau(i)=b$. Then is it true that $\sigma$ and $\tau$ must have different signs (i.e., one is even and the other is odd)?

Context: This problem arose when I was trying to verify the proof that a determinant function on $R$ given by $det(\alpha_{ij})=\sum_{\sigma\in S_n}\epsilon(\sigma)\alpha_{\sigma(1)1}\dots\alpha_{\sigma(n)n}$ is indeed a $n$-multilinear alternating form.

My Attempt: I have tried a few different cases, and the statement seems to be true (which is what I wish to see); but to show it rigorously proves to be trickier. I am currently trying to make use of $\tau=\sigma\sigma^{-1}\tau$.

Answer 1

The sign of a permutation is given by

$$\operatorname{sgn}(\sigma)=(-1)^{N(\sigma)}$$

where $N(\sigma)$ is the number of inversions.

As the transposition consists of adjacent elements, the number of inversions changes by exactly one, hence the sign changes parity.

Answer 2

Note $\sigma^{-1}\tau(x)=x \forall x\in[n]-\{i, i+1\}$, $\sigma^{-1}\tau(i)=i+1$, $\sigma^{-1}\tau(i+1)=i$. Thus $\sigma^{-1}\tau$ is the transposition $(i\ i+1)$. Thus $\tau=\sigma(i\ i+1)$.