# 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 $G$.

## 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 $H = K$.