Homomorph-containing not implies no nontrivial homomorphism to quotient group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., homomorph-containing subgroup) need not satisfy the second subgroup property (i.e., normal subgroup having no nontrivial homomorphism to its quotient group)
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Statement

It is possible to have a group $G$ and a subgroup $H$ such that:

1. $H$ is a homomorph-containing subgroup in $G$, i.e., for any homomorphism in $\mathrm{H}\mathrm{o}\mathrm{m}(H,G)$ the image of $H$ is contained in $H$.
2. $H$ is not a normal subgroup having no nontrivial homomorphism to its quotient group. In other words, there exists a nontrivial homomorphism from $H$ to $G/H$.

Proof

Take the following:

• $G$ is cyclic group:Z4.
• $H$ is the subgroup Z2 in Z4 (hence isomorphic to cyclic group:Z2), which is the unique subgroup generated by elements of order two.

Then:

• $H$ is homomorph-containing in $G$: This is because the image of $H$ under any homomorphism must also have exponent at most two, hence must be inside $H$.
• There is a nontrivial homomorphism from $H$ to $G/H$: In fact, both are isomorphic to cyclic group:Z2.