Complete divisibility-closedness is strongly intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., completely divisibility-closed subgroup) satisfying a subgroup metaproperty (i.e., strongly intersection-closed subgroup property)
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Statement

Suppose $G$ is a group and $H_{i},i\in I$ are all completely divisibility-closed subgroups of $G$ . Then, the intersection of subgroups $H=\bigcap _{i\in I}H_{i}$ is also completely divisibility-closed.

Here, a subgroup is completely divisibility-closed if for any prime number $p$ such that every element of the group has a $p^{th}$ root in the group, all $p^{th}$ roots of any element in the subgroup are in the subgroup.

Related facts

Proof

Given: A group $G$ , completely divisibility-closed subgroups $H_{i},i\in I$ of $G$ . A prime number $p$ such that $G$ is $p$ -divisible. An element $g\in H=\bigcap _{i\in I}H_{i}$ . An element $x\in G$ such that $x^{p}=g$ .

To prove: $x\in H$

Proof: It suffices to demonstrate the last sentence, because the existence of $p^{th}$ roots in $G$ is guaranteed by $G$ being $p$ -divisible.

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	$g\in H_{i}$ for each $i\in I$ .	$H=\bigcap _{i\in I}H_{i}$ , $g\in H$ .		direct from given
2	$x\in H_{i}$ for each $i\in I$ .	$x^{p}=g$ , $H_{i}$ is completely divisibility-closed.	Step (1)	Step-given direct.
3	$x\in H$ .	$H=\bigcap _{i\in I}H_{i}$	Step (2)	Step-given direct