Powering-invariance is union-closed

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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

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.