Directed union of subgroups is subgroup

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 $$G$$.

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.