Difference between revisions of "Index is multiplicative"
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+ | ==Statement== | ||
+ | |||
+ | ===Set-theoretic version=== | ||
+ | |||
+ | Suppose <math>K \le H \le G</math> are groups. Then, we have a natural surjective map between the [[left coset space]]s: | ||
+ | |||
+ | <math>G/K \to G/H</math> | ||
+ | |||
+ | with the property that the inverse image of each point has size equals to the size of the left coset space <math>H/K</math>. | ||
+ | |||
+ | ===Numerical version=== | ||
+ | |||
+ | Suppose <math>K \le H \le G</math> are groups such that the [[fact about::index of a subgroup|indices]] <math>[G:H]</math> and <math>[H:K]</math> are finite. Then, we have: | ||
+ | |||
+ | <math>[G:K] = [G:H][H:K]</math>. | ||
+ | |||
+ | 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 accordingly. | ||
+ | |||
+ | ==Related facts== | ||
+ | |||
+ | * [[Lagrange's theorem]]: Lagrange's theorem is a special case of this where <math>K</math> 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 <math>G/K</math> to <math>G/H</math> is a homomorphism and the kernel is <math>H/K</math>. | ||
+ | |||
+ | ==Facts used== | ||
+ | |||
+ | # [[uses::Subgroup containment implies coset containment]]: If <math>K \le H \le G</math> ,then every left coset of <math>K</math> is contained in a unique left coset of <math>H</math>. | ||
+ | |||
+ | ==Proof== | ||
+ | |||
+ | ===Proof of the set-theoretic version=== | ||
+ | |||
+ | '''Given''': Groups <math>K \le H \le G</math>. | ||
+ | |||
+ | '''To prove''': There is a surjective map from <math>G/K</math> to <math>G/H</math> where the inverse image of every point has size equal to the size of <math>H/K</math>. | ||
+ | |||
+ | '''Proof''': Define <math>\varphi</math> as the map sending a coset of <math>K</math> to the unique coset of <math>H</math> containing it (fact (1)). In other words: | ||
+ | |||
+ | <math>\varphi(gK) = gH</math>. | ||
+ | |||
+ | <math>\varphi</math> is a well-defined map from <math>G/K</math> to <math>G/H</math>. Further: | ||
+ | |||
+ | * <math>\varphi</math> is surjective, since for any coset <math>gH</math> of <math>H</math> in <math>G</math>, <math>gH = \varphi(gK)</math>. | ||
+ | * The size of each inverse image equals the size of <math>H/K</math>: Consider a coset <math>gK</math>. We want to find all the left cosets of <math>H</math> which map to this. This is equivalent to finding all the left cosets of <math>H</math> contained in <math>gK</math>. | ||
+ | |||
+ | For this, consider a map <math>\psi</math> that sends a left coset <math>xH</math> of <math>H</math> in <math>gK</math>, to the left coset <math>g^{-1}xH</math>. | ||
+ | |||
+ | Note that: | ||
+ | |||
+ | * This map <math>\psi</math> is well-defined, because if <math>x</math> is replaced by <math>xh, h \in H</math>, the left cosets <math>g^{-1}xH</math> and <math>g^{-1}xhH</math> are the same. | ||
+ | * The map <math>\psi</math> sends left cosets of <math>H</math> in <math>gK</math> to left cosets of <math>H</math> in <math>K</math>, because since <math>x \in gK</math>, <math>g^{-1}x \in K</math>. | ||
+ | * The map is injective, because it comes from a left multiplication on cosets. | ||
+ | * The map is surjective, because any left coset <math>kH</math> of <math>H</math> in <math>K</math> arises as the image of the left coset <math>gkH</math>, which lies in <math>gK</math>. | ||
+ | |||
+ | Thus, <math>\psi</math> is a bijection between the left cosets of <math>H</math> in <math>gK</math> and the left cosets of <math>H</math> in <math>K</math>. Thus, the number of left cosets of <math>H</math> in <math>gK</math> equals the size of the left coset space <math>K/H</math>. | ||
+ | |||
+ | ===Proof of the numerical version=== | ||
+ | |||
+ | The set-theoretic version shows that <math>G/K</math> is the disjoint union of <math>|G/H|</math> sets, each of size <math>|H/K|</math>. This yields: | ||
+ | |||
+ | <math>|G/K| = |G/H||H/K|</math>. | ||
+ | |||
+ | By the definition of index of a subgroup, this yields: | ||
+ | |||
+ | <math>[G:K] = [G:H][H:K]</math>. | ||
+ | |||
==References== | ==References== | ||
Revision as of 12:21, 16 October 2008
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:
.
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 accordingly.
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 . Further:
- is surjective, since for any coset of in , .
- The size of each inverse image equals the size of : Consider a coset . We want to find all the left cosets of which map to this. This is equivalent to finding all the left cosets of contained in .
For this, consider a map that sends a left coset of in , to the left coset .
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)