Abelian-quotient not implies kernel of a bihomomorphism

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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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about abelian-quotient subgroup|Get more facts about kernel of a bihomomorphism
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property abelian-quotient subgroup but not kernel of a bihomomorphism|View examples of subgroups satisfying property abelian-quotient subgroup and kernel of a bihomomorphism

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.