Quotient-powering-invariance is union-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., quotient-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 quotient-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 quotient-powering-invariant subgroup of $G$.

Related facts

• Powering-invariance is union-closed

Facts used

1. Normality is strongly join-closed

Proof

Note that existence of $p^{th}$ roots is guaranteed (see divisibility is inherited by quotient groups). The part we need to establish is uniqueness. Note also that by Fact (1), the union must be a normal subgroup if it is a subgroup, hence we will assume this in our setup.

Given: $G$ is a group and $H_{i},i\in I$ is a collection of quotient-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 $p$. Two elements $g_{1},g_{2}\in G$ that are in the same coset of $H$ (concretely, $g_{1}g_{2}^{-1}\in H$). $x_{1},x_{2}\in G$ are respectively the unique solutions to $x_{1}^{p}=g_{1}$ and $x_{2}^{p}=g_{2}$.

To prove: $x_{1}$ and $x_{2}$ are in the same coset of $H$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists $i\in I$ such that $g_{1}g_{2}^{-1}\in H_{i}$. In other words, there exists $i\in I$ such that $g_{1}$ and $g_{2}$ are in the same coset of $H_{i}$.		$H$ is the union of $H_{i},i\in I$, $g_{1}g_{2}^{-1}\in H$.		Follows directly from given.
2	$x_{1}$ and $x_{2}$ are in the same coset of $H_{i}$.		$G$ is $p$-powered, $H_{i}$ is quotient-powering-invariant in $G$.	Step (1)	Direct from given and Step (1). Note that quotient-powering-invariant in a $p$-powered group means that the $p^{th}$ roots of elements in the same coset are in the same coset.
3	$x_{1}$ and $x_{2}$ are in the same coset of $H$.		$H$ is the union of $H_{i},i\in I$.	Step (2)	Since $H_{i}\leq H$, being in the same coset of $H_{i}$ implies being in the same coset of $H$.