Kernel of a bihomomorphism implies completely divisibility-closed

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., kernel of a bihomomorphism) must also satisfy the second subgroup property (i.e., completely divisibility-closed subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about kernel of a bihomomorphism|Get more facts about completely divisibility-closed subgroup

Suppose $G$ is a group and $H$ is a subgroup of $G$. Suppose $H$ is a kernel of a bihomomorphism in $G$, i.e., there exists a bihomomorphism:

$b:G\times G\to M$

for some group $M$, such that:

$H=\{x\in G\mid b(x,y){\mbox{ is the identity element of }}M{\mbox{ for all }}y\in G\}$

Then, $H$ is a completely divisibility-closed subgroup of $G$ and hence also a completely divisibility-closed normal subgroup of $G$.

Facts used

1. Annihilator of divisibility-closed subgroup under bihomomorphism is completely divisibility-closed

Proof

The proof follows directly from Fact (1), using the fact that the divisibility-closed subgroup in question here is the whole group.