Index is multiplicative
This article gives the statement, and possibly proof, of a subgroup property (i.e., subgroup of finite index) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Contents
Statement
Set-theoretic version
Suppose are groups. Then, we have a natural surjective map between the left coset spaces:
with the property that the inverse image of each point has size equals to the size of the left coset space .
Numerical version
Suppose are groups such that the indices and are finite. Then, we have:
.
In particular, if has finite index in and has finite index in , then has finite index in .
Note also that if is finite, then so are the other two indices, and we thus get:
.
Note that the statement makes sense even for infinite groups, if we interpret the cardinalities of the coset spaces as infinite cardinals and do the multiplication using the rules for product of cardinals.
Related facts
- Lagrange's theorem: Lagrange's theorem is a special case of this where is the trivial group.
- Third isomorphism theorem: The third isomorphism theorem is a stronger version of the statement where both subgroups are normal in the whole group. In this case, the surjective map from to is a homomorphism and the kernel is .
Facts used
- Subgroup containment implies coset containment: If ,then every left coset of is contained in a unique left coset of .
Proof
Proof of the set-theoretic version
Given: Groups .
To prove: There is a surjective map from to where the inverse image of every point has size equal to the size of .
Proof: Define as the map sending a coset of to the unique coset of containing it (fact (1)). In other words:
.
is a well-defined map from to by Fact (1).
Surjectivity of the map
is surjective, since for any coset of in , .
Size of the inverse images
To prove': Consider the map . Choose and consider the left coset . Then, the subset of comprising left cosets that map to under can be put in bijection with .
Proof: We explicitly construct such a bijection based on the choice of , defined as:
Note that:
- This map is well-defined, because if is replaced by , the left cosets and are the same.
- The map sends left cosets of in to left cosets of in , because since , .
- The map is injective, because it comes from a left multiplication on cosets.
- The map is surjective, because any left coset of in arises as the image of the left coset , which lies in .
Thus, is a bijection between the left cosets of in and the left cosets of in . Thus, the number of left cosets of in equals the size of the left coset space .
Proof of the numerical version
The set-theoretic version shows that is the disjoint union of sets, each of size . This yields:
.
By the definition of index of a subgroup, this yields:
.
References
Textbook references
- Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, ^{More info}, Page 77, Exercise 4 of Miscellaneous Problems (asked only for a finite group)