The union of a chain of subgroups is a subgroup

Let G be a group, and \{ H_i \}_{i \in \mathbb{Z}} be an ascending chain of subgroups of G; that is, H_i \subseteq H_j for i \leq j. Prove that \bigcup_{i \in \mathbb{Z}} H_i is a subgroup of G.


Note that \bigcup_{i \in \mathbb{Z}} H_i is not empty since 1 \in \bigcup_{i \in \mathbb{Z}} H_i. Now let x,y \in \bigcup_{i \in \mathbb{Z}} H_i. Then we have x \in H_a and y \in H_b for some a,b \in \mathbb{Z}; suppose without loss of generality that a \leq b. Then H_a \subseteq H_b, so that x \in H_b. By the subgroup criterion, then, xy^{-1} \in H_b, so that xy^{-1} \in \bigcup_{i \in \mathbb{Z}} H_i. By the subgroup criterion, \bigcup_{i \in \mathbb{Z}} H_i is a subgroup of G.

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