Difference between revisions of "Normal equals potentially characteristic"

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(Proof of (1) implies (2) (hard direction))
(Proof of (1) implies (2) (hard direction))
 
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| 4 || <math>V</math> is characteristic in <math>K</math>. || || || Steps (2), (3) || Under any automorphism of <math>K</math>, the image of <math>V</math> is a homomorphic image of <math>V</math> in <math>K</math>. Its projection to <math>K/V \cong G</math> is a homomorphic image of <math>V</math> in <math>G</math>, which is trivial by step (3), so the image of <math>V</math> in <math>K</math> must be in <math>V</math>.
 
| 4 || <math>V</math> is characteristic in <math>K</math>. || || || Steps (2), (3) || Under any automorphism of <math>K</math>, the image of <math>V</math> is a homomorphic image of <math>V</math> in <math>K</math>. Its projection to <math>K/V \cong G</math> is a homomorphic image of <math>V</math> in <math>G</math>, which is trivial by step (3), so the image of <math>V</math> in <math>K</math> must be in <math>V</math>.
 
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| 5 || The centralizer of <math>V</math> in <math>VH</math> equals <math>H</math>. || || || Steps (1), (2) || By definition of the wreath product action, <math>H</math> centralizes <math>V</math>. Since <math>S</math> is centerless, <math>V</math> is also centerless. Thus, <math>C_{VH}(V)</math> contains <math>H</math> but has trivial intersection with <math>V</math>, forcing <math>C_{VH} (V)= H</math>.
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| 5 || The centralizer of <math>V</math> in <math>VH</math> equals <math>H</math>. || || || Steps (1), (2) || By definition of the wreath product action, <math>H</math> centralizes <math>V</math>. Since <math>S</math> is centerless, <math>V</math> is also centerless. Thus, <math>C_{VH}(V)</math> contains <math>H</math> but has trivial intersection with <math>V</math>, forcing <math>C_{VH} (V) = H</math>.
 
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| 6 || The centralizer of <math>V</math> in <math>K</math> equals <math>H</math>. || || || Steps (2), (5) || Step (5) already shows that <math>C_{VH}(V) = H</math>, so it suffices to show that <math>C_K(V) \le VH</math>. To see this, note that, by the construction in step (2), any element of <math>K</math> outside <math>VH</math> permutes the direct factors of <math>V</math> as an element of <math>G</math> outside <math>H</math>. This action is nontrivial, so the action is nontrivial, and hence elements outside <math>VH</math> cannot centralize <math>V</math>. This forces <math>C_K(V) \le VH</math>, completing the proof.
 
| 6 || The centralizer of <math>V</math> in <math>K</math> equals <math>H</math>. || || || Steps (2), (5) || Step (5) already shows that <math>C_{VH}(V) = H</math>, so it suffices to show that <math>C_K(V) \le VH</math>. To see this, note that, by the construction in step (2), any element of <math>K</math> outside <math>VH</math> permutes the direct factors of <math>V</math> as an element of <math>G</math> outside <math>H</math>. This action is nontrivial, so the action is nontrivial, and hence elements outside <math>VH</math> cannot centralize <math>V</math>. This forces <math>C_K(V) \le VH</math>, completing the proof.

Latest revision as of 17:03, 10 July 2019

This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a subgroup H of a group G :

  1. H is a normal subgroup of G.
  2. H is a potentially characteristic subgroup of G in the following sense: there exists a group K containing G such that H is a characteristic subgroup of K.

Related facts

Stronger facts

Other related facts

Facts used

  1. Characteristicity is centralizer-closed
  2. Characteristic implies normal
  3. Normality satisfies intermediate subgroup condition

Proof

Proof of (1) implies (2) (hard direction)

Given: A group G, a normal subgroup H of G.

To prove: There exists a group K containing G such that H is characteristic in K.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let S be a simple non-abelian group that is not isomorphic to any subgroup of G. Note that such a group exists. For instance, we can take the finitary alternating group on any set of cardinality strictly bigger than that of G.
2 Let K be the restricted wreath product of S and G, where G acts via the regular action of G/H and let V be the restricted direct power S^{G/H}. In other words, K is the semidirect product of the restricted direct power V = S^{G/H} and G, acting via the regular group action of G/H. Step (1)
3 Any homomorphism from V to G is trivial. Steps (1), (2) By definition, V is a restricted direct product of copies of S. Since S is simple and not isomorphic to any subgroup of G, any homomorphism from S to G is trivial. Thus, any homomorphism from V to G is trivial.
4 V is characteristic in K. Steps (2), (3) Under any automorphism of K, the image of V is a homomorphic image of V in K. Its projection to K/V \cong G is a homomorphic image of V in G, which is trivial by step (3), so the image of V in K must be in V.
5 The centralizer of V in VH equals H. Steps (1), (2) By definition of the wreath product action, H centralizes V. Since S is centerless, V is also centerless. Thus, C_{VH}(V) contains H but has trivial intersection with V, forcing C_{VH} (V) = H.
6 The centralizer of V in K equals H. Steps (2), (5) Step (5) already shows that C_{VH}(V) = H, so it suffices to show that C_K(V) \le VH. To see this, note that, by the construction in step (2), any element of K outside VH permutes the direct factors of V as an element of G outside H. This action is nontrivial, so the action is nontrivial, and hence elements outside VH cannot centralize V. This forces C_K(V) \le VH, completing the proof.
7 H is characteristic in K. Fact (1) Steps (4), (6) Step-fact combination direct.
This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Proof of (2) implies (1) (easy direction)

Given: A group G, a subgroup H of G, a group K containing G such that H is characteristic in K.

To prove: H is normal in G.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 H is normal in K. Fact (2) H is characteristic in K -- Given-fact-combination direct.
2 H is normal in G. Fact (3) H \le G \le K Step (1) Given-step-fact combination direct.