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 $⟨H_i{⟩}_{i\in I}$. 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 $H_i\le H$, being in the same coset of $H_i$ implies being in the same coset of $H$.