Exercise 4.6.5 (Chain of subgroups $G_1 \leq G_2 \leq \cdots \leq G$ such that $G = \bigcup_{i=1}^\infty G_i$ and each $G_i$ is simple)

Proof. By hypothesis, there exists a chain $G_1\le G_2\le \cdots \le G_k\le \cdots$ of subgroup of $G$ such that $G={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{G}_i$, where each $G_i$ is simple.

Let $H\ne \{1\}$ be a normal subgroup of $G$. We want to show $H=G$.

For every index $i$, $H\cap G_i⊴G_i$, where $G_i$ is simple, thus

If for all $i$, $H\cap G_i=\{1\}$, then

which is excluded. Therefore there is some index $k$ such that

Then $H\cap G_k=G_k$, so $G_k\le H$.

We prove by induction that $G_i\le H$ for all $i\ge k$.

Suppose that $G_i\le H$ for some positive integer $i\ge k$. Then $H\cap G_{i+1}⊴G_{i+1}$. Since $G_{i+1}$ is simple, $H\cap G_{i+1}=\{1\}$ or $H\cap G_{i+1}=G_{i+1}$. But $\{1\}\ne H\cap G_k\le H\cap G_{i+1}$, thus $H\cap G_{i+1}\ne \{1\}$. This shows that $H\cap G_{i+1}=G_{i+1}$, so $G_{i+1}\le H$.

The induction is done, which proves for all $i\ge k$,

Therefore, since $G_1\le G_2\le \cdots \le G_k$,

so $H=G$. This proves that $G$ is simple.

If there exists a chain of subgroups of $G$, $G_1\le G_2\le \cdots \le G_k\le \cdots$, such that $G={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{G}_i$ and each $G_i$ is simple then $G$ is simple. □