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 $G$.

Related facts

• Quotient-powering-invariance is union-closed
• Powering-invariance is not finite-join-closed

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.