Isomorph-freeness is strongly join-closed

Verbal statement
Suppose $$G$$ is a group, and $$H_i, i \in I$$ is a collection of isomorph-free subgroups of $$G$$ for some (possibly empty) indexing set $$I$$. Then, the join of the $$H_i$$ is also an isomorph-free subgroup of $$G$$.

Related facts

 * Intermediate isomorph-conjugacy is normalizing join-closed
 * Intermediate automorph-conjugacy is normalizing join-closed

Proof
Given: A group $$G$$, a collection $$H_i, i \in I$$ of isomorph-free subgroups of $$G$$ for some (possibly empty) indexing set $$I$$.

To prove: The join of the $$H_i$$s is also isomorph-free.

Proof: Suppose $$H$$ is the join of the $$H_i$$s, and suppose $$K$$ is a subgroup of $$G$$ isomorphic to $$H$$. Let $$\alpha:H \to K$$ be an isomorphism, and let $$K_i = \alpha(H_i)$$. Now, since $$\alpha$$ is an isomorphism, $$H_i \cong K_i$$ for each $$i \in I$$. By assumption, $$H_i$$ is isomorph-free in $$G$$, so $$H_i = K_i$$ for each $$i \in I$$. Thus, the join of the $$H_i$$s equals the join of the $$K_i$$s, forcing $$H = K$$.