The union of a chain of simple groups is simple

Prove that if there exists a chain of subgroups G_1 \leq G_2 \leq \cdots \leq G such that G = \bigcup_{i \in \mathbb{Z}} G_i and each G_i is simple then G is simple.

Suppose we have such a chain, and let N \leq G be a normal subgroup. Note that N \cap G_i \leq G_i is a normal subgroup for each G_i. Since each G_i is simple, we have N \cap G_i \in \{ 1, G_i \}.

Suppose N \cap G_i = 1 for all G_i. Then N = N \cap G = N \cap \bigcup_{i \in \mathbb{Z}} G_i = \bigcup_{i \in \mathbb{Z}} (N \cap G_i) = \bigcup_{i \in \mathbb{Z}} 1 = 1. Hence N = 1.

Suppose now that N \cap G_k = G_k for some k. Note that if N \cap G_i = G_i for some i, then G_i \leq  N \cap G_{i+1} = G_{i+1}. Thus if we let k be minimal such that N \cap G_k = G_k, then by induction N \cap G_i = G_i for all i \geq k. Then N = N \cap G = N \cap \bigcup_{i \in \mathbb{Z}} G_i = N \cap \left( \bigcup_{i=0}^{k-1} G_i \cup \bigcup_{i=k}^\infty G_i \right) = \bigcup_{i=0}^{k-1}(N \cap G_i) \cup \bigcup_{i=k}^\infty (N \cap G_i) = \bigcup_{i=0}^{k-1} 1 \cup \bigcup_{i=k}^\infty G_i = 1 \cup G = G. Thus N = G.

Since every normal subgroup of G is trivial, G is simple.

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