# 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.