Abelian characteristic is not join-closed

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

It is possible to have a group $G$ with Abelian characteristic subgroups (i.e., subgroups that are both Abelian and characteristic) $H$ and $K$ such that the join $\langle H,K\rangle$ is not an Abelian characteristic subgroup.

Since the join is always a characteristic subgroup, the particular thing that can fail is that the join need not be Abelian.

Related facts

• Characteristicity is strongly join-closed
• Abelian normal is not join-closed
• Cyclic normal is not join-closed

Proof

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