Intermediate isomorph-conjugacy is normalizing join-closed

Statement
Suppose $$H, K$$ are intermediately isomorph-conjugate subgroups of a group $$G$$, such that $$K \le N_G(H)$$. Then, the join $$HK$$ (which is also equal to the product of subgroups) is also intermediately isomorph-conjugate.

Related facts

 * Isomorph-freeness is strongly join-closed
 * Intermediate automorph-conjugacy is normalizing join-closed
 * Pronormality is normalizing join-closed

Proof
This proof follows the left-action convention, though it can be stated equally well using the right-action convention (a similar proof for pronormality uses the right-action convention -- the proofs use essentially the same idea). We denote by $$c_g$$ the operation $$x \mapsto gxg^{-1}$$ of conjugation by $$g$$.

Given: A group $$G$$, intermediately isomorph-conjugate subgroups $$H,K \le G$$ such that $$K$$ normalizes $$H$$. $$\sigma$$ is an isomorphism from $$HK$$ to some subgroup of $$G$$.

To prove: There exists $$l \in \langle HK, \sigma(HK) \rangle$$ such that $$c_l(HK) = \sigma(HK)$$.

Proof:


 * 1) (Given data used: $$H$$ is intermediately isomorph-conjugate in $$G$$): $$H$$ and $$\sigma(H)$$ are isomorphic subgroups, so, since $$H$$ is intermediately isomorph-conjugate, there exists a $$g \in \langle H, \sigma(H) \rangle$$ such that $$c_g(H) := gHg^{-1} = \sigma(H)$$. Note that $$g \in \langle H, \sigma(H) \rangle \le \langle HK, \sigma(HK) \rangle$$.
 * 2) (No given data used): Define $$\sigma' = c_g^{-1} \circ \sigma$$ as a map from $$HK$$ to $$G$$. Now, $$\sigma'(HK) \le \langle HK, \sigma(HK) \rangle$$, because it involves composing $$\sigma$$ with conjugation by an element inside $$\langle H, \sigma(H) \rangle \le \langle HK, \sigma(HK) \rangle$$. Thus, $$\langle HK, \sigma'(HK) \rangle \le \langle HK, \sigma(HK) \rangle$$.
 * 3) (Given data used: $$K$$ normalizes $$H$$): We have $$K \le N_G(H)$$. Now, $$\sigma'(H) = H$$ by construction. Thus, $$\sigma'(K) \le N_G(\sigma'(H)) = N_G(H)$$. Thus, $$\langle K, \sigma'(K) \rangle \le N_G(H)$$.
 * 4) (Given data used: $$K$$ is intermediately isomorph-conjugate in $$G$$): Since $$K$$ is intermediately isomorph-conjugate in $$G$$, the conclusion of step (3) tells us that there exists $$k \in \langle K, \sigma'(K) \rangle$$ such that $$c_k(K) = \sigma'(K)$$. In particular, $$k \in N_G(H)$$, so $$c_k(H) = H$$. Thus, $$c_k(HK) = c_k(H)c_k(K) = H\sigma'(K) = \sigma'(H)\sigma'(K) = \sigma'(HK)$$, with $$k \in \langle K, \sigma'(K) \le \langle HK, \sigma'(HK)\rangle$$. By the conclusion of step (2), this yields $$k \in \langle HK, \sigma(HK) \rangle$$.
 * 5) Now, consider the product $$l = gk$$ We claim that this works (No problem data used): Clearly, since both $$g$$ and $$k$$ are elements of $$\langle HK, \sigma(HK) \rangle$$, so is $$l$$. Further, $$c_l = c_g \circ c_k$$. In particular, $$c_l(HK) = c_g(c_k(HK)) = c_g(\sigma'(HK)) = \sigma(HK)$$. Thus, $$HK$$ and $$\sigma(HK)$$ are conjugate in the subgroup they generate.