Difference between revisions of "Index is multiplicative"

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{{subgroup metaproperty satisfaction|
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property = subgroup of finite index|
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metaproperty = transitive subgroup property}}
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==Statement==
 
==Statement==
  
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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:
 
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:
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<math>[G:K] = [G:H][H:K]</math>.
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In particular, if <math>K</math> has finite index in <math>H</math> and <math>H</math> has finite index in <math>G</math>, then <math>K</math> has finite index in <math>G</math>.
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Note also that if <math>[G:K]</math> is finite, then so are the other two indices, and we thus get:
  
 
<math>[G:K] = [G:H][H:K]</math>.
 
<math>[G:K] = [G:H][H:K]</math>.

Revision as of 13:23, 1 November 2008

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)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about subgroup of finite index |Get facts that use property satisfaction of subgroup of finite index | Get facts that use property satisfaction of subgroup of finite index|Get more facts about transitive subgroup property


Statement

Set-theoretic version

Suppose K \le H \le G are groups. Then, we have a natural surjective map between the left coset spaces:

G/K \to G/H

with the property that the inverse image of each point has size equals to the size of the left coset space H/K.

Numerical version

Suppose K \le H \le G are groups such that the indices [G:H] and [H:K] are finite. Then, we have:

[G:K] = [G:H][H:K].

In particular, if K has finite index in H and H has finite index in G, then K has finite index in G.

Note also that if [G:K] is finite, then so are the other two indices, and we thus get:

[G:K] = [G:H][H:K].

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 K 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 G/K to G/H is a homomorphism and the kernel is H/K.

Facts used

  1. Subgroup containment implies coset containment: If K \le H \le G ,then every left coset of K is contained in a unique left coset of H.

Proof

Proof of the set-theoretic version

Given: Groups K \le H \le G.

To prove: There is a surjective map from G/K to G/H where the inverse image of every point has size equal to the size of H/K.

Proof: Define \varphi as the map sending a coset of K to the unique coset of H containing it (fact (1)). In other words:

\varphi(gK) = gH.

\varphi is a well-defined map from G/K to G/H. Further:

  • \varphi is surjective, since for any coset gH of H in G, gH = \varphi(gK).
  • The size of each inverse image equals the size of H/K: Consider a coset gK. We want to find all the left cosets of H which map to this. This is equivalent to finding all the left cosets of H contained in gK.

For this, consider a map \psi that sends a left coset xH of H in gK, to the left coset g^{-1}xH.

Note that:

  • This map \psi is well-defined, because if x is replaced by xh, h \in H, the left cosets g^{-1}xH and g^{-1}xhH are the same.
  • The map \psi sends left cosets of H in gK to left cosets of H in K, because since x \in gK, g^{-1}x \in K.
  • The map is injective, because it comes from a left multiplication on cosets.
  • The map is surjective, because any left coset kH of H in K arises as the image of the left coset gkH, which lies in gK.

Thus, \psi is a bijection between the left cosets of H in gK and the left cosets of H in K. Thus, the number of left cosets of H in gK equals the size of the left coset space K/H.

Proof of the numerical version

The set-theoretic version shows that G/K is the disjoint union of |G/H| sets, each of size |H/K|. This yields:

|G/K| = |G/H||H/K|.

By the definition of index of a subgroup, this yields:

[G:K] = [G:H][H:K].

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)