Homomorph-containment is strongly join-closed

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

Statement with symbols

Suppose G is a group, I is an indexing set, and H_i, i \in I, is a collection of homomorph-containing subgroups of G. Then, the join of subgroups \langle H_i \rangle_{i \in I} is also a homomorph-containing subgroup of G.

Related facts

Related facts about join-closed

Proof

Given: A group G, an indexing set I, a collection H_i of homomorph-containing subgroups of G, i \in I. H = \langle H_i \rangle_{i \in I}. A homomorphism \varphi: H \to G.

To prove: \varphi(H) is containined in H.

Proof: \varphi(H) = \varphi \langle H_i \rangle_{i \in I} = \langle \varphi(H_i) \rangle_{i \in I}. Since each H_i is homomorph-containing in G, \varphi(H_i) is contained in H_i, so the join of \varphi(H_i), i \in I, is contained in the join of the H_is, which is H.