Isomorph-freeness is strongly join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., isomorph-free subgroup) satisfying a subgroup metaproperty (i.e., strongly join-closed subgroup property)
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Statement

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 $H=K$.