Exercise 2.1.15 (Union of an ascending chain of subgroups)

Proof. Let $H_1\le H_2\le \cdots$ be an ascending chain of subgroups of $G$ (with identity $e=1_G$). Put

• Since $H_1$ is a subgroup of $G$, $e\in H_1$. Therefore $e\in H={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{H}_i,$ so $H\ne \varnothing$.
• Let $x,y$ be any elements of $H$. Then there are indices $i,j$ such that $x\in H_i$ and $y\in H_j$. Put $k=\max <!--\; FUNCTION\; APPLICATION\; -->(i,j)$, so $k\ge i,k\ge j$. Since $H_1\subseteq H_2\subseteq \cdots$, then $x\in H_k$ and $y\in H_k$, thus $x{y}^{-1}\in H_k$. This shows that $x{y}^{-1}\in H={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{H}_i.$

Therefore $H$ is a subgroup of $G$.

If $H_1\le H_2\le \cdots$ is an ascending chain of subgroups of $G$, ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{H}_i$ is a subgroup of $G$. □