Subhomomorph-containment is transitive

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

• Homomorph-containment is not transitive
• Subhomomorph-containing implies right-transitively homomorph-containing
• Full invariance is transitive

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.