Abelian-quotient not implies kernel of a bihomomorphism

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., abelian-quotient subgroup) need not satisfy the second subgroup property (i.e., kernel of a bihomomorphism)
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Statement

There are three different but equivalent formulations (in the sense that any example for one formulation also furnishes examples for the other):

1. It is possible to have an abelian group $G$ such that there is no bihomomorphism $b:G\times G\to M$ to any group $M$ with kernel the trivial subgroup of $G$.
2. It is possible to have a group $G$ and an abelian-quotient subgroup $H$ of $G$ such that there is no bihomomorphism $G\times G\to M$ for any group $M$ with kernel $H$.
3. It is possible to have a nilpotent group $G$ such that there is no bihomomorphism $b:G\times G\to M$ to any group $M$ with kernel the [[derived subgroup] of $G$.

Facts used

1. Kernel of a bihomomorphism implies completely divisibility-closed

Proof

We will furnish an example for (1), which also gives examples for (2) and (3).

Let $p$ be a prime number. Let $G$ be the quasicyclic group for $p$, i.e., the group of all $(p^{n})^{th}$ roots of unity for all $n$. $G$ is an abelian group.

The trivial subgroup is not completely divisibility-closed in $G$: $G$ is $p$-divisible but the identity element has $p^{th}$ roots outside the trivial subgroup.