Conjugacy-closedness is not upper join-closed

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

We can have the following situation: $H\leq G$ is a subgroup, $K_{1},K_{2}$ are intermediate subgroups of $G$ containing $H$, such that $H$ is conjugacy-closed in $K_{1}$ as well as in $K_{2}$, but not in the join $\langle K_{1},K_{2}$.

Proof

A generic example

Suppose $H$ is a group with an automorphism $\sigma$ of order two that is not class-preserving: the automorphism does not preserve conjugacy classes. For instance, $H$ could be an Abelian group of exponent greater than two, and $\sigma$ could be the inverse map.

Let $A$ be the subgroup of $\operatorname {Aut} (H\times H)$ generated by the automorphism $(\sigma ,\sigma )$ (i.e., $\sigma$ acting coordinate-wise) and the coordinate exchange automorphism. Since both these automorphisms are of order two and commute, $A$ is a Klein-four group. Define $G=(H\times H)\rtimes A$ with the specified action.

Now, let $B_{1}$ be the two-element subgroup of $A$ generated by the coordinate exchange automorphism, and $B_{2}$ be the two-element subgroup of $A$ generated by the composite the coordinate exchange automorphism and $(\sigma ,\sigma )$. Define $K_{1}=(H\times H)B_{1}$ and $K_{2}=(H\times H)B_{2}$.

• The subgroup $H=H\times \{e\}$ is conjugacy-closed in $K_{1}$: Any element of $K_{1}$ is a product of an element in $H\times H$ and an element in $H\times H$. Note that the element of $H\times H$ preserves conjugacy classes, so it remains to see the effect of the element of $B_{1}$. if the element of $B_{1}$ is trivial, it of course acts as the identity. If it is not trivial, it sends every element of $H\times \{e\}$ to an element of $\{e\}\times H$, so no two distinct conjugacy classes in $H\times \{e\}$ get fused by the action of $B_{1}$.
• The subgroup $H=H\times \{e\}$ is conjugacy-closed in $K_{1}$: The reasoning is identical to the above reasoning.
• The subgroup $H=H\times \{e\}$ is not conjugacy-closed in $G$: Indeed, the automorphism $(\sigma ,\sigma )$ does not preserve conjugacy classes in $H\times \{e\}$.

A similar kind of example can be constructed when the automorphism $\sigma$ is not class-preserving, and has finite order $n>2$. In this case, we need to case a $n$-fold direct product of $H$, and have the diagonal automorphism $\sigma$ as well as the cyclic coordinate permutation automorphism act on this direct product.

Some particular examples

The smallest particular example of the above is when $H$ is the cyclic group of order three, and $\sigma$ is the inverse map. $G$ in this case has order $36$, and $K_{1},K_{2}$ both have order $18$.

Within nilpotent groups, the smallest particular example is when $H$ is the cyclic group of order four, and $\sigma$ is the inverse map. $G$ has order $64$, and $K_{1},K_{2}$ both have order $32$.