Directed union of subgroups is subgroup

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This article gives the statement, and possibly proof, of a basic fact in group theory.
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Statement

Verbal statement

The union of a nonempty directed set of subgroups of a group is again a subgroup.

Statement with symbols

Suppose G is a group, I a nonempty directed set, and H_i, i \in I is a collection of subgroups of G indexed by I, such that i \le j \implies H_i \le H_j. Then, the subset of G given by:

\bigcup_{i \in I} H_i

is also a subgroup of G.

Definitions used

Directed set

A partially ordered set I is termed directed if for any i,j \in I, there exists k \in I, such that i \le k, j \le k.

Proof

Given: G is a group, I a nonempty directed set, and H_i, i \in I is a collection of subgroups of G indexed by I, such that i < j \implies H_i \le H_j.


To prove: The subset of G given by:

\bigcup_{i \in I} H_i

is also a subgroup of G.

Proof: We check the three conditions for a subgroup:

  1. Identity element: Indeed, the identity element of G is in all the H_is, so it is in their union.
  2. Inverse elements: Suppose x is in the union. Then, x \in H_i for some i \in I. Thus, x^{-1} \in H_i (because H_i is a subgroup). So, x^{-1} is in the union.
  3. Products: Suppose x,y are in the union. Then, x \in H_i, y \in H_j for some i,j \in I. By the directedness property, there exists k \in I, such that i \le k, j \le k. Thus, H_i \le H_k and H_j \le H_k. In particular, both x and y are in the subgroup H_k. So, their product xy is in H_k, so xy is in the union.