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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 $G$.


Solution: 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 $G$.

Linearity

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