# Infinite Simple Group as Union of Solvable Groups

Let $A_n$ denote the alternating group on first $n$ natural numbers, and $A_{\mathbb{N}}$ be the union $\cup_{n\geq 1} \,\, A_n$. (In other words, $A_{\mathbb{N}}$ is the set of all bijections from $\mathbb{N}$ to $\mathbb{N}$ which move only finitely many points, and they are even permutations on these finitely many points.)

Question: Is there an infinite ascending chain $1\leq H_1\leq H_2\leq \cdots$ of finite solvable subgroups of $A_{\mathbb{N}}$ such that $\cup_{n\geq 1}=A_{\mathbb{N}}$?

No, there isn't. If there were such a chain, then for a large enough $n$ we would have $A_5 \leq H_n$. Since $H_n$ is solvable, it would also mean that $A_5$ is solvable (solvability is preserved when taking subgroups). But $A_5$ is simple non-abelian, so this is a contradiction.
To put the same in other words: $A_{\mathbb{N}}$ is not locally solvable, and any union of an ascending chain of solvable subgroups must be locally solvable.