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Transfer cycle decomposition formula
Statement
Let be a finite group, , an abelian group written multiplicatively, a homomorphism, and the transfer of Transfer homomorphism for a finite index subgroup, which is independent of the transversal by Transfer is independent of the transversal. Let and let
be the decomposition of the finite set of right cosets into the orbits of the right multiplication action of the cyclic subgroup (The subgroup generated by a subset, the cyclic subgroup , and cyclic groups, , and every cyclic group is abelian). Put and choose with . Then
Facts & Assumptions
Given: A finite group , a subgroup , an abelian group , a homomorphism , an element , and the transfer of Transfer homomorphism for a finite index subgroup.
For the coset is defined for every , the assignment is a right action of on the finite set , each factor () lies in , and products of finitely many elements of the abelian group are independent of the order of the factors (Transfer homomorphism for a finite index subgroup).
The transfer does not depend on the transversal (Transfer is independent of the transversal).
If satisfy then , and conversely ( iff , and iff ).
is a subgroup of , and for all (, and every cyclic group is abelian, Exponent laws in a group: and for all , and when and commute).
The orbits of the action of the subgroup on partition (The orbits of a group action are the equivalence classes of iff for some , and hence partition the acted-on set, The orbit and stabilizer of a point in a group action, Left group actions, transitive actions, and faithful actions).
Every nonempty set of positive integers has a least element, and for integers and there are with and (The well-ordering principle, Division with remainder in : for and there are unique with and ).
Proof
Fix . Since is finite, the elements cannot all be distinct, so for some ; applying the inverse permutation (which is the action of ) gives , an equality of the form with . By [F7] there is a least positive integer with .
For every one has if and only if : if then by [F5] and step 1.1 (including , since as well), while for arbitrary writing with by [F7] gives , so minimality forces .
The orbit of under is , and by step 2.1 it equals , whose elements are pairwise distinct. In particular and .
The orbit decomposition of [F6] is therefore a decomposition into finitely many sets each of the form with and ; write .
The rule for defines a transversal of : indeed every coset of lies in exactly one and hence equals exactly one , and represents it, because by [F1].
Because is independent of the transversal by [F2], it may be computed with the transversal of step 5.1: , where and .
For one has , so the corresponding factor equals by [F4].
For one has by step 4.1, so and the corresponding factor equals , an element of because by step 4.1 and [F3].
Multiplying the contributions of steps 7.1 and 7.2 over all and , all factors with are and the remaining one for each is ; hence . ∎
Depends on
- Transfer homomorphism for a finite index subgroup
- Transfer is independent of the transversal
- Left and right cosets $gH$ and $Hg$ of a subgroup
- $x\in aH$ iff $a^{-1}x\in H$, and $aH=bH$ iff $a^{-1}b\in H$
- A group homomorphism automatically satisfies $f(e) = e'$ and $f(g^{-1}) = f(g)^{-1}$, and $f(g^{n}) = f(g)^{n}$ for every $n \in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed
- Division with remainder in $\mathbb{Z}$: for $a \in \mathbb{Z}$ and $b > 0$ there are unique $q, r \in \mathbb{Z}$ with $a = qb + r$ and $0 \le r < b$
- The well-ordering principle
- The orbits of a group action are the equivalence classes of $x\sim y$ iff $y=g\cdot x$ for some $g$, and hence partition the acted-on set
- Left group actions, transitive actions, and faithful actions
- The orbit $G\cdot x$ and stabilizer $G_x$ of a point in a group action
- The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups
- $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian
- Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ **when $g$ and $h$ commute**
Used by
Dependency tree · two levels
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Sources
- Paul Flavell, An Introduction to Transfer and Fusion in Finite Groups, §§2–5 (standard reference, not scraped)
- Hans Kurzweil and Bernd Stellmacher, The Theory of Finite Groups, §§7.1–7.2 (standard reference, not scraped)