Normal equals potentially characteristic

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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:

  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.
  3. Any homomorphism from V to G is trivial: 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, for any homomorphism from V to G is trivial.
  4. V is characteristic in K: 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, so the image of V in K must be in V.
  5. The centralizer of V in K equals H: By definition, H centralizes V. Using the fact that S is centerless and that inner automorphisms of S cannot be equal to conjugation by elements in G \setminus H, we can show that it is precisely the center.
  6. H is characteristic in K: This follows from the previous two steps and fact (1).

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.