CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles

Statement

A cycle of length $k$ has sign $(- 1)^{k - 1}$ . If $σ \in S_{n}$ and $c (σ)$ is the number of cycles after every fixed point is included as a one-cycle, then

$sgn (σ) = (- 1)^{n - c (σ)} .$

Facts & Assumptions

Given: A natural $n$ and a permutation $σ \in S_{n}$ .

[L1]

Sign is a homomorphism, every finite permutation has a disjoint-cycle decomposition, and a $k$ -cycle is a product of $k - 1$ transpositions (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ , Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation, Every finite permutation is a product of transpositions, so the transpositions generate $S_{n}$ ).

Proof

technique · direct

1.1givenL1

The standard factorisation of a $k$ -cycle has $k - 1$ transpositions, so [L1] gives sign $(- 1)^{k - 1}$ .

2.1step 1.1L1

Write the disjoint-cycle decomposition of $σ$ with lengths $k_{1}, \dots, k_{r}$ . Multiplicativity of sign and step 1.1 give $sgn (σ) = (- 1)^{\sum_{i} (k_{i} - 1)}$ .

3.1step 2.1L1∎

Insert each fixed point as a one-cycle. Then the cycle lengths sum to $n$ , the number of cycles is $c (σ)$ , and $\sum_{i} (k_{i} - 1) = n - c (σ)$ , which gives the formula.

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, Group Theory, Corollary 4.27

A kk-cycle has sign (−1)k−1(−1)k−1, and sgn⁡(σ)=(−1)n−c(σ)sgn(σ)=(−1)n−c(σ) when fixed points are counted as cycles