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