Product formula
This article describes a natural isomorphism between two structures or between a family of structures
Contents
Statement
Set-theoretic version
Suppose are subgroups. Then, there is a natural bijection between the left coset spaces:
.
Note that while , being an intersection of subgroups, is a subgroup, is not necessarily a subgroup.
Numerical version
Let and be two subgroups of a finite group . Then:
Here is the product of subgroups.
Related facts
- Second isomorphism theorem: This is a stronger formulation of the set-theoretic version, which holds when both the groups in the denominator are normal in the respective numerators. In this case, the natural bijection turns out to be an isomorphism.
Corollaries
- Index of intersection of permuting subgroups divides product of indices: If , then the index of divides the product of the index of and the index of .
- Index satisfies transfer inequality: This states that if , then .
- Index satisfies intersection inequality: This states that if are subgroups, then .
Facts used
- Subgroup containment implies coset containment: If are subgroups, then every left coset of is contained in a left coset of .
- Lagrange's theorem
Proof
Proof of the set-theoretic version
Given: A group , and subgroups .
To prove: There is a natural bijection between the coset spaces and .
Proof: We first define the map:
as follows:
.
In other words, it sends each coset of to the coset of containing it.
- The map sends cosets to cosets: Note first that if two elements are in the same coset of , they are in the same coset of . Thus, the map sends cosets of to cosets of . (This is fact (1)).
- The map is well-defined with the specified domain and co-domain: Further, if , then . In other words, if the original coset is in , the new coset is in . Thus, the map is well-defined from to .
- The map is injective: Finally, . That means that , forcing . But we anyway have , so , forcing that and are in the same coset of . Thus, .
- The map is surjective: Any left coset of in can be written as where . Thus, we can write where . Then, , with . Thus, .
Proof of the numerical version using the set-theoretic version
The numerical version follows by combining the set-theoretic version with Lagrange's theorem:
.
Notice that the left side measures the number of cosets of in . Since all these cosets are disjoint and have size equal to , the left side is . Similarly, the right side is . This yields:
which, upon rearrangement, gives the product formula.
(Note: Although for the left side, we can quote Lagrange's theorem to say that , we cannot directly quote Lagrange's theorem for the right side, because is not necessarily a group. However, the reason is precisely the same in both cases: is a union of left cosets of , each having the same size as , so the number of such cosets is .)