Isomorph-freeness is strongly join-closed

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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

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_is is also isomorph-free.

Proof: Suppose H is the join of the H_is, 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_is equals the join of the K_is, forcing H = K.