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: $H \le G$ 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 $32$.