Kernel of a bihomomorphism implies completely divisibility-closed

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:

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

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

Facts used

 * 1) uses::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.