Show that union of ascending chain of subgroups is subgroup

[Mr Davis 97]

Homework Statement

Let $H_1 \le H_2 \le \cdots$ be an ascending chain of subgroups of $G$. Prove that $H = \bigcup\limits_{i=1}^{\infty} H_{i}$ is a subgroup of $G$.

Homework Equations

The Attempt at a Solution

Certainly $H$ is nonempty, since each subgroup $H_i$ has at least the identity element. Now, let $a,b \in H$. Then $a \in H_i$ where $i$ is taken to be minimal. Also $b \in H_j$, where $j$ is taken to be minimal. WLOG suppose that $i \le j$. Then $H_i \subseteq H_j$ and so $a,b \in H_j$. Then since $H_j$ is a subgroup, $ab^{-1} \in H_j \subseteq H$, and so $ab^{-1} \in H$.

 

[fresh_42]

Homework Statement

Let $H_1 \le H_2 \le \cdots$ be an ascending chain of subgroups of $G$. Prove that $H = \bigcup\limits_{i=1}^{\infty} H_{i}$ is a subgroup of $G$.

Homework Equations

The Attempt at a Solution

Certainly $H$ is nonempty, since each subgroup $H_i$ has at least the identity element. Now, let $a,b \in H$. Then $a \in H_i$ where $i$ is taken to be minimal. Also $b \in H_j$, where $j$ is taken to be minimal. WLOG suppose that $i \le j$. Then $H_i \subseteq H_j$ and so $a,b \in H_j$. Then since $H_j$ is a subgroup, $ab^{-1} \in H_j \subseteq H$, and so $ab^{-1} \in H$.

Correct.