# 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 $G$.

## 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, $g_1g_2^{-1} \in H$). $x_1,x_2 \in G$ 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 $g_1g_2^{-1} \in H_i$. 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 $H_i, i \in I$, $g_1g_2^{-1} \in H$. 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 $H$.