Subhomomorph-containment is transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subhomomorph-containing subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Statement

Suppose H \le K \le G. Suppose K is a subhomomorph-containing subgroup of G and H is a subhomomorph-containing group of K. Then, H is a subhomomorph-containing subgroup of G.

Related facts

Proof

Given: Groups H \le K \le G. A homomorphism \varphi: A \to G for a subgroup A of H.

To prove: \varphi(A) is a subgroup of H.

Proof: Since A is a subgroup of H, A is a subgroup of K. Thus, \varphi:A \to G is a homomorphism from a subgroup of K. Since K is subhomomorph-containing in G, \varphi(A) is contained in K. Thus, \varphi can be viewed as a map from A to K.

Thus, \varphi:A \to K is a map from a subgroup A of H. Since H is subhomomorph-containing in K, \varphi(A) \le H, and we are done.