# Isomorph-freeness is strongly join-closed

From Groupprops

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 is a group, and is a collection of isomorph-free subgroups of for some (possibly empty) indexing set . Then, the join of the is also an isomorph-free subgroup of .

## Related facts

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

## Proof

**Given**: A group , a collection of isomorph-free subgroups of for some (possibly empty) indexing set .

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

**Proof**: Suppose is the join of the s, and suppose is a subgroup of isomorphic to . Let be an isomorphism, and let . Now, since is an isomorphism, for each . By assumption, is isomorph-free in , so for each . Thus, the join of the s equals the join of the s, forcing .