## 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.