No nontrivial homomorphism to quotient group not implies complemented normal

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., normal subgroup having no nontrivial homomorphism to its quotient group) need not satisfy the second subgroup property (i.e., complemented normal subgroup)
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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 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 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 $G$.