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A -cycle has sign , and when fixed points are counted as cycles
Statement
A cycle of length has sign . If and is the number of cycles after every fixed point is included as a one-cycle, then
Facts & Assumptions
Given: A natural and a permutation .
Sign is a homomorphism, every finite permutation has a disjoint-cycle decomposition, and a -cycle is a product of transpositions (The sign is a homomorphism , surjective exactly when , 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 ).
Proof
The standard factorisation of a -cycle has transpositions, so [L1] gives sign .
Write the disjoint-cycle decomposition of with lengths . Multiplicativity of sign and step 1.1 give .
Insert each fixed point as a one-cycle. Then the cycle lengths sum to , the number of cycles is , and , which gives the formula.
Depends on
Used by
- A five-cycle is even and a six-cycle is odd Example
- A₄ consists of the identity, eight 3-cycles, and three products of disjoint transpositions Example
- The conjugacy classes of A₅: sizes 1,20,15,12,12 and the split 5-cycles Example
- The five conjugacy classes of S₄ and the class equation 24=1+6+3+8+6 Example
- The seven conjugacy classes of S₅ and their centralizer and class sizes Example
- V₄={1,(12)(34),(13)(24),(14)(23)} is a proper nontrivial normal subgroup of A₄ Example
- FALSE: Aₙ is simple for every n≥4 False statement
- FALSE: two even permutations of the same cycle type are always conjugate in Aₙ False statement
- Every nontrivial normal subgroup of Aₙ contains a 3-cycle for n≥5 Lemma
- Aₙ is generated by 3-cycles for every n≥3 Theorem
- Aₙ is simple for every n≥5 Theorem
- Every group of order 30 has normal Sylow 3- and 5-subgroups and is not simple Theorem
- For n≥2, an Sₙ-class of an even permutation splits in Aₙ exactly when all cycle lengths, including 1-cycles, are odd and distinct Theorem
- The cycle index of Aₙ is the parity-filtered symmetric-group sum Theorem
Dependency tree · two levels
15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- J. S. Milne, Group Theory, Corollary 4.27 (standard reference, not scraped)