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)
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
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 .
We will furnish an example for (1), which also gives examples for (2) and (3).
The trivial subgroup is not completely divisibility-closed in : is -divisible but the identity element has roots outside the trivial subgroup.