# Powering-invariance is union-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) satisfying a subgroup metaproperty (i.e., union-closed subgroup property)
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## Statement

Suppose $G$ is a group and $H_i,i \in I$ is a collection of powering-invariant subgroups of $G$. Suppose the union $\bigcup_{i \in I} H_i$ is a subgroup of $G$ (and hence is also the same as the join of subgroups $\langle H_i \rangle_{i \in I}$). Then, this union of also a powering-invariant subgroup of $G$.

## Proof

Given: $G$ is a group and $H_i,i \in I$ is a collection of powering-invariant subgroups of $G$. The union $H = \bigcup_{i \in I} H_i$ is a subgroup of $G$. $G$ is powered over a prime $p$. An element $g \in H$.

To prove: There exists $x \in H$ such that $x^p = g$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists $i \in I$ such that $g \in H_i$. $H$ is the union of $H_i, i \in I$, $g \in H$. Follows directly from given.
2 $H_i$ is $p$-powered. $G$ is $p$-powered, $H_i$ is powering-invariant in $G$. Directly from given data used.
3 There exists $x \in H_i$ such that $x^p = g$. Steps (1), (2) direct from steps.
4 There exists $x \in H$ such that $x^p = g$. $H$ is the union of $H_i, i \in I$ Step (3) Step-given direct.