# 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.

## 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 Facts used 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