Chain of simple subgroups (D&F 4.6.5)

Prove that if there exists a chain of subgroups \(G_1 \le G_2 \le \ldots \le G\) such that \(G = \cup_{i=1}^\infty G_i\) and each \(G_i\) is simple then \(G\) is simple.

Suppose \(H \trianglelefteq G\) and \(H \neq 1,G\); so \(H \cap G_i \trianglelefteq G_i\) for all \(i\). We want to show that there is some \(i\) such that \(H \cap G_i\) is a proper nontrivial subgroup of \(G_i\), which contradicts the fact that \(G_i\) is simple. What follows is simply accounting for all the possibilities:

1. \(H \cap G_i = G_i\) for all \(i\)
In this case, we have \(G_i \le H\) for all \(i\) and so \(G = \cup_{i=1}^\infty G_i \le H\), a contradiction.

2. \(H \cap G_i = 1\) for all \(i\)
It follows that \(H \cap G = H \cap (\cup_{i=1}^\infty G_i) = 1\), yet another contradiction.

3. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i < j\)
The second statement implies that \(G_i \le G_j \le H\), contradicting the first statement.

4. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i > j\)
The first statement implies that \(H \cap G_j \le H \cap G_i = 1\), contradicting the second statement.