Homomorph-containing not implies no nontrivial homomorphism to quotient group

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 $$\operatorname{Hom}(H,G)$$ 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 particular example::cyclic group:Z4.
 * $$H$$ is the subgroup particular example::Z2 in Z4 (hence isomorphic to particular example::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.