Directed union of subgroups is subgroup

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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:

Identity element: Indeed, the identity element of $G$ is in all the $H i$ s, so it is in their union.
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.
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 $x y$ is in $H k$ , so $x y$ is in the union.