Characteristicity is not upper join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property).
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Statement

Property-theoretic statement

The subgroup property of being a characteristic subgroup does not satisfy the subgroup metaproperty of being upper join-closed.

Statement with symbols

It is possible to find a group G, a subgroup H of G, and intermediate subgroups K,L of G containing H, such that H is characteristic in both K and L, but H is not characteristic in \langle K, L \rangle.

Proof

A generic example

Let H be a group and A,B be subgroups of H such that:

  • \langle A, B \rangle = H
  • There is no nontrivial homomorphism from H to A
  • There is no nontrivial homomorphism from H to B

Then, set G = H \times H, K = H \times A, L = H \times B. We then observe that:

  • H is a characteristic subgroup inside K (in fact, H is fully characteristic inside K).
  • H is a characteristic subgroup inside L (in fact, H is fully characteristic inside L).
  • H is not characteristic inside G = \langle K, L \rangle.

Note that this generic example also shows the the property of being a fully characteristic subgroup is not closed under upper joins. For full proof, refer: Full characteristicity is not upper join-closed

Some specific examples

One example is to take H as a simple group, and A, B as proper subgroups generating it. For instance, H is the alternating group on five letters, and A,B can be taken, for instance, as two of the embedded alternating groups on four letters.

Another generic example

Let M be a non-Abelian group generated by Abelian subgroups A,B. Consider:

  • G = M \times M
  • H = M' \times \{ e \}
  • K = M \times A
  • L = M \times B.

Then, H is the commutator subgroup both of K and of L, so H is characteristic in both. On the other hand, H is not characteristic in G because the coordinate exchange automorphism (that swaps the two copies of M) does not preserve H.

This generic example also shows that the property of being fully characteristic is not preserved under upper joins. For full proof, refer: Full characteristicity is not upper join-closed

An example is M a non-Abelian group of order p^3, and A and B two Abelian subgroups that generate it.