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

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 $h$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists $y\in H$ such that $y^{p}=h$.
2	$(b(x,y))^{p}=b(g,y)=b(x,h)$.		$b$ is a bihomomorphism		Since $b$ is a bihomomorphism, $(b(x,y))^{p}=b(x^{p},y)=b(g,y)$ (using the fact that it is a homomorphism in the first coordinate). Also, $(b(x,y))^{p}=b(x,y^{p})=b(x,h)$ (using the fact that it is a homomorphism in the second coordinate).
3	$b(g,y)$ is the identity element of $M$.		$g\in \operatorname {Ann} _{G}(H)$.	Step (1)	By Step (1), $y\in H$. Combined with the given, we get that $b(g,y)$ is the identity element of $M$.
4	$b(x,h)$ is the identity element of $M$.			Steps (2), (3)	Step-combination direct

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.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists $x\in G$ such that $x^{p}=g$.		$G$ is $p$-divisible.
2	For every $x\in G$ such such that $x^{p}=g$, $x\in \operatorname {Ann} _{G}(H)$.	The "version for a particular prime"	$G$ is $p$-divisible, $H$ is divisibility-closed in $G$.		From the given, $H$ is also $p$-divisible, so we can apply the "version for a particular prime" and get the conclusion.