Characteristicity is not upper join-closed

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.