No nontrivial homomorphism to quotient group not implies complemented normal

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ such that $$H$$ is a normal subgroup having no nontrivial homomorphism to its quotient group (i.e., the set $$\operatorname{Hom}(H,G/H)$$ is a singleton set comprising the trivial homomorphism) but $$H$$ is not a complemented normal subgroup in $$G$$.

Related facts

 * No common composition factor with quotient group not implies complemented
 * Normal Hall implies permutably complemented

Proof
We take the following:


 * $$G$$ is the particular example::binary octahedral group (of order 48). This is the double cover of symmetric group:S4 of "-" type.
 * $$H$$ is the unique subgroup of $$G$$ isomorphic to particular example::special linear group:SL(2,3) (order 24, quotient group is cyclic group:Z2).

This example works because:


 * There is no nontrivial homomorphism from $$H$$ to $$G/H$$: Indeed, as per the subgroup structure of special linear group:SL(2,3), $$H$$ has no subgroup of index two.
 * $$H$$ is not a complemented normal subgroup in $$G$$: As per the subgroup structure of binary octahedral group, the only subgroup of order two in $$G$$ is the center of binary octahedral group, which is inside $$H$$. Hence, $$H$$ has no complement in $$G$$.