How statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
FALSE: two even permutations of the same cycle type are always conjugate in
Statement refuted
Two elements of with the same cycle type must be conjugate in .
Facts & Assumptions
Given: The cycles and in .
A -cycle has sign (A -cycle has sign , and when fixed points are counted as cycles), and is the kernel of sign (The alternating group of even permutations).
Conjugation relabels cycle entries (Conjugating a cycle relabels each entry: ).
For and , the -class of splits into two -classes of equal size exactly when all cycle lengths in its decomposition, including -cycles for fixed points, are odd and no two are equal (For , an -class of an even permutation splits in exactly when all cycle lengths, including -cycles, are odd and distinct).
Counterexample
In , the cycles and have sign by [F1] and have the same cycle type .
The lengths and are odd and distinct, so [F3] says that the -class of -cycles splits into two -classes.
More explicitly, [F2] shows that conjugating to reverses the cyclic order on while fixing the remaining point; every such relabeling is odd. Hence no element of performs it.
Thus the two displayed even permutations have the same cycle type but are not conjugate in .
Depends on
- For $n\ge2$, an $S_n$-class of an even permutation splits in $A_n$ exactly when all cycle lengths, including $1$-cycles, are odd and distinct
- Conjugating a cycle relabels each entry: $g(a_1\,\ldots\,a_k)g^{-1}=(g(a_1)\,\ldots\,g(a_k))$
- The alternating group $A_n=\ker(\operatorname{sgn})$ of even permutations
- A $k$-cycle has sign $(-1)^{k-1}$, and $\operatorname{sgn}(\sigma)=(-1)^{n-c(\sigma)}$ when fixed points are counted as cycles
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 34 results over 11 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)