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

Statement
Suppose $$G$$ and $$M$$ are groups and:

$$b: G \times G \to M$$

is a bihomomorphism. Suppose $$H$$ is a subgroup of $$G$$. Define:

$$\operatorname{Ann}_G(H) := \{ g \in G \mid b(g,h) \mbox{ is the identity element of } M \ \forall \ h \in H \}$$

Version for a particular prime
If $$p$$ is a prime number such that $$H$$ is $$p$$-divisible, then for any element of $$\operatorname{Ann}_G(H)$$, all $$p^{th}$$ roots of that element in $$G$$ must be in $$\operatorname{Ann}_G(H)$$.

Corollary for divisibility-closed
In particular, if $$H$$ is a divisibility-closed subgroup of $$G$$, $$\operatorname{Ann}_G(H)$$ is a completely divisibility-closed subgroup of $$G$$.

Proof of version for a particular prime
Given: A bihomomorphism $$b: G \times G \to M$$ of groups. A subgroup $$H$$ of $$G$$. A prime number $$p$$ such that $$H$$ is $$p$$-divisible.

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

An element $$g \in \operatorname{Ann}_G(H)$$, and an element $$x \in G$$ such that $$x^p = g$$.

To prove: $$x \in \operatorname{Ann}_G(H)$$.

Proof: Pick an arbitrary element $$h \in H$$. It will suffice to show that $$b(x,h)$$ is the identity element of $$M$$, independent of the choice of $$h$$.

Proof of corollary for divisibility-closed
Assume that we have already established the version for each particular prime.

Given: A bihomomorphism $$b: G \times G \to M$$ of groups. A divisibility-closed subgroup $$H$$ of $$G$$. A prime number $$p$$ such that $$G$$ is $$p$$-divisible.

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

An element $$g \in \operatorname{Ann}_G(H)$$.

To prove: There exists an element of $$\operatorname{Ann}_G(H)$$ whose $$p^{th}$$ power is $$G$$, and every element of $$G$$ whose $$p^{th}$$ power is $$G$$ is in $$\operatorname{Ann}_G(H)$$.

Proof: We break the proof in two steps.