# Abelian-quotient not implies kernel of a bihomomorphism

From Groupprops

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) neednotsatisfy 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):

- It is possible to have an abelian group such that there is
*no*bihomomorphism to any group with kernel the trivial subgroup of . - It is possible to have a group and an abelian-quotient subgroup of such that there is
*no*bihomomorphism for any group with kernel . - It is possible to have a nilpotent group such that there is
*no*bihomomorphism to any group with kernel the [[derived subgroup] of .

## Facts used

## Proof

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

Let be a prime number. Let be the quasicyclic group for , i.e., the group of all roots of unity for all . is an abelian group.

The trivial subgroup is *not* completely divisibility-closed in : is -divisible but the identity element has roots outside the trivial subgroup.