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:

1. Identity element: Indeed, the identity element of $G$ is in all the $H_i$s, 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.