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
.)