Characteristicity is not upper join-closed

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 $$\langle K, L \rangle$$.

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.

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.

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