Quotient-powering-invariance is union-closed

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) uses::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: