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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.