# The sign of a permutation is well-defined

The symmetric group \(S_n\) contains elements which are made up from transpositions (proof). It is a fact that if \(\sigma \in S_n\) can be made by multiplying together an even number of transpositions, then it cannot be made by multiplying an odd number of transpositions, and vice versa.

# Proof

Let \(c(\sigma)\) be the number of cycles in the disjoint cycle decomposition of \(\sigma \in S_n\), including singletons. For example, \(c\) applied to the identity yields \(n\), while \(c((12)) = n-1\) because \((12)\) is shorthand for \((12)(3)(4)\dots(n-1)(n)\). We claim that multiplying \(\sigma\) by a transposition either increases \(c(\sigma)\) by \(1\), or decreases it by \(1\).

Indeed, let \(\tau = (kl)\). Either \(k, l\) lie in the same cycle in \(\sigma\), or they lie in different ones.

If they lie in the same cycle, then \($\sigma = \alpha (k a_1 a_2 \dots a_r l a_s \dots a_t) \beta\)$ where \(\alpha, \beta\) are disjoint from the central cycle (and hence commute with \((kl)\)). Then \(\sigma (kl) = \alpha (k a_s \dots a_t)(l a_1 \dots a_r) \beta\), so we have broken one cycle into two.

If they lie in different cycles, then \($\sigma = \alpha (k a_1 a_2 \dots a_r)(l b_1 \dots b_s) \beta\)$ where again \(\alpha, \beta\) are disjoint from \((kl)\). Then \(\sigma (kl) = \alpha (k b_1 b_2 \dots b_s l a_1 \dots a_r) \beta\), so we have joined two cycles into one.

Therefore \(c\) takes even values if there are evenly many transpositions in \(\sigma\), and odd values if there are odd-many transpositions in \(\sigma\).

More formally, let \(\sigma = \alpha_1 \dots \alpha_a = \beta_1 \dots \beta_b\), where \(\alpha_i, \beta_j\) are transpositions.

Parents:

- Symmetric group
The symmetric groups form the fundamental link between group theory and the notion of symmetry.

Just fyi, as stated the sentence “Therefore \(c(\sigma)\) takes even values if there are evenly many transpositions in \(\sigma\), and odd values if there are odd-many transpositions in \(c(\sigma)\) is false as stated. It depends on the parity of \(n\) --- indeed, \(c({\rm identity}) = n\).