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Two elements of are conjugate if and only if they have the same cycle type
Statement
For , there is a with if and only if and have the same cycle type, including their numbers of fixed points.
Facts & Assumptions
Given: Permutations .
Every permutation has a disjoint-cycle decomposition, unique up to cycle order, cyclic rotation, and omission of fixed points (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation).
Cycle type records the number of cycles of each length , including -cycles (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).
Conjugating a cycle by relabels all its entries by (Conjugating a cycle relabels each entry: ).
Proof
Suppose . Conjugate every factor in the disjoint-cycle decomposition of ; [F3] preserves each cycle length and relabels fixed points.
Conversely, suppose the cycle types agree. By [F1] and [F2], pair every cycle of with a cycle of of the same length, including the fixed points, and define by sending entries positionwise in each pair.
Hence and have the same multiplicities and the same cycle type.
The paired cycles partition the underlying set, so the map in step 1.2 is a bijection and hence an element of .
Applying [F3] to each paired cycle gives .
Depends on
- Conjugating a cycle relabels each entry: $g(a_1\,\ldots\,a_k)g^{-1}=(g(a_1)\,\ldots\,g(a_k))$
- Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation
- Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 34 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- K. Conrad, Conjugacy Classes (standard reference, not scraped)