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 $C$ be a group and $A,B$ be subgroups of $C$ such that:

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

Then, set $H=C\times \{e\}$ (here, $\{e\}$ denotes the trivial group), $G=C\times C,K=C\times A,L=C\times B$. We then observe that:

• $H$ is a characteristic subgroup inside $K$ (in fact, $H$ is fully invariant inside $K$).
• $H$ is a characteristic subgroup inside $L$ (in fact, $H$ is fully invariant inside $L$).
• $H$ is not characteristic inside $G=\langle K,L\rangle$, since it is not invariant under the automorphism exchanging the two coordinates in the direct product.

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 invariance is not upper join-closed

Some specific examples

One example is to take $C$ as a simple non-abelian group, and $A,B$ as proper subgroups generating it. For instance, $C$ 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 derived 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 invariant is not preserved under upper joins. For full proof, refer: Full invariance 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.