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 $i\leq j\implies H_{i}\leq H_{j}$. 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 $i\leq k,j\leq k$.

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 $i<j\implies H_{i}\leq H_{j}$.

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 $i\leq k,j\leq k$. Thus, $H_{i}\leq H_{k}$ and $H_{j}\leq H_{k}$. 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.