Find an Explicit Isomorphism Between $S^1$ and a Subgroup of $\text{Aut}(\mathbb{N})$

Question

This is a followup question to this Math Stack Question where it was shown that $S^1\cong G\leq\text{Aut}(\mathbb{N})$ for some subgroup $G$ of $\text{Aut}(\mathbb{N})$(although I admittedly don't follow the whole discussion). I wanted to answer the question by constructing an isomorphism, but failed along the way. I want to see if my method would work, or find some other explicit isomorphism. Here is my thought process. $$$$ Consider the sequence of the first $n$ primes $(p_1=2,p_2=3,p_3,\ldots,p_n)$ and let $N=\prod_{i=1}^n p_i$. Now consider the group under addition $\mod(2\pi)$ (which by the given construction is a subgroup of $S^1$)(note that $\frac{2\pi}{4}$, for instance, is not in this set): $$\left\{ 2\pi\left(\sum_{i=1}^{n} \frac{a_i}{p_i}\right): a_i\in\mathbb{Z} \right\}\equiv_{2\pi} \left\{ 2\pi\frac{a}{N}: a\in\mathbb{Z}, 0\leq a<N\right\}$$ Which is isomorphic to some subgroup of $S_N$, $G_n$. So here is where I am not sure how to tackle this problem, can we show in some sense that $\lim_{n\to\infty}G_n$ converges to some group $G$ and that $G\cong [\{2\pi a/p: p \text{ is a product of finitely many distinct primes, and } a\in\mathbb{Z}\}\mod(2\pi)]\leq S^1$? If this is the case, can we further conclude that there is some form of completion of $G$, in the sense of 'Cauchy' infinite compositions of permutations, which is isomorphic to $S^1$?

$$$$ If you have some other explicit isomorphism which actually works, please do share! $$$$ EDIT: It would appear that there is a much more straightforward approach. Consider the cyclic permutation in $\text{Perm}(\mathbb{N})$: $$\sigma_{a+1} = (a(a+1)/2,\ldots,(a+1)(a+2)/2-1)$$ for $a\in \mathbb{N}$ and $\sigma_1=id$. Consider the subgroup $A$ of $\text{Perm}(\mathbb{N})$ where $\sigma\in A$ if $$\sigma = \circ_{i=1}^{\infty}(\sigma_{b_i})^{c_i}$$ for some convergent series $$\sum_{i=1}^{\infty}\frac{c_i}{b_i}$$ where $c:\mathbb{N}\longrightarrow \mathbb{Z}$ and $b:\mathbb{N}\longrightarrow \mathbb{N}$. Let $B<A$ be the subgroup such that $\sigma\in A$ is in $B$ if the corresponding Cauchy sequence is congruent to $0$ modulo $1$. I think it seems intuitive that $A/B\cong S^1$.

EDIT2: I will first engage in a discussion about sequences of permutations, then attempt to prove the claim that $A/B\cong S^1$. First, we will define convergence of a sequence of permutations as follows: $\sigma:\mathbb{N}\longrightarrow \text{Perm}(\mathbb{N})$ converges to $\pi\in \text{Perm}(\mathbb{N})$ if $$(\forall M)(\exists N)(\forall n>N)(\forall 1\leq m<M)\text{ }\sigma_{n}(m) = \pi(m)$$ Next, we will say that a sequence of permutations has the 'Cauchy Property' if $$(\forall M)(\exists N)(\forall n,p>N)(\forall 1\leq m<M)\text{ }\sigma_{p}^{-1}\circ\sigma_{n}(m) = m$$ We can quite easily demonstrate equivalence: \begin{align*} (\implies): &\text{ let $(\sigma_n)$ converge to $\pi$, then}\\ &(\forall M)(\exists N)(\forall n,p>N)(\forall 1\leq m<M)\sigma_{p}^{-1}\circ\sigma_{n}(m) = \pi^{-1}\circ\pi(m) = m.\\ (\impliedby): &\text{ let $(\sigma_n)$ have the Cauchy Property, but suppose $(\sigma_n)$ does not converge, then}\\ &(\forall \pi)(\exists M)(\forall N)(\exists n>N)(\exists 1\leq m<M)\sigma_n(m)\neq \pi(m) \text{ so}\\ &(\exists M)(\forall N)(\exists n,p>N)(\exists 1\leq m<M)\sigma_n(m)\neq \sigma_p(m)\\ &\text{ So $(\sigma_n)$ does not have the Cauchy Property, which is a contradiction.} \end{align*}

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Now we will demonstrate that sequences of partial compositions of elements in $A$ have the Cauchy Property. From here on out we are reserving the permutation $\sigma_n$ as notation for the cycle: $$\sigma_{a+1} = (a(a+1)/2,\ldots,(a+1)(a+2)/2-1)$$ for $a\in \mathbb{N}$ and $\sigma_1=id$. Let $\sigma \in A$ where $\sigma = \circ_{i=1}^{\infty}(\sigma_{b_i})^{c_i} = \lim_{n\rightarrow\infty}\circ_{i=1}^{n}(\sigma_{b_i})^{c_i}$, let $\pi_n = \circ_{i=1}^{n}(\sigma_{b_i})^{c_i}$. It is given that $$\sum_{i=1}^{\infty}\frac{c_i}{b_i}$$ converges (in $\mathbb{R}$) where $c:\mathbb{N}\longrightarrow \mathbb{Z}$ and $b:\mathbb{N}\longrightarrow \mathbb{N}$. Thus $(c_i/b_i)\to 0$ so either $(\exists N)(\forall i\geq N)$, $c_i=0$ (in which case $\pi_n\to \pi_N$) or $$(\forall \varepsilon=(1/M) >0)(\exists N)(\forall i>N)b_i>\frac{|c_i|}{\varepsilon}\geq\frac{1}{\varepsilon}>M \text{ or }c_i = 0$$ Thus $(\forall M'=M(M+1)/2)(\exists N)(\forall n,p>N)(\forall 1\leq m<M')$ \begin{align*} \pi_p^{-1}\circ \pi_n(m) &=[\circ_{i=1}^{p}(\sigma_{b_i})^{c_i}]^{-1}\circ [\circ_{i=1}^{n}(\sigma_{b_i})^{c_i}](m)\\ &= [\circ_{i=1}^{N}(\sigma_{b_i})^{c_i}]^{-1}\circ [\circ_{i=1}^{N}(\sigma_{b_i})^{c_i}](m) \text{(using $N$ such that $b_i > M'$, or $c_i=0$ when $i>N$)}\\ &= m \end{align*} So for every $\sigma\in A$, the associated sequence of partial compositions has the Cauchy Property and thus converges. Next we will show that composition in $A$ converges and equals component wise ccomposition. Let $\kappa, \kappa' \in A$, $\kappa= \lim_{n\to\infty}\pi_n = \lim_{n\to\infty}\circ_{i=1}^{n}(\sigma_{b_i})^{c_i}$ and $\kappa'= \lim_{n\to\infty}\pi'_n = \lim_{n\to\infty}\circ_{i=1}^{n}(\sigma_{b'_i})^{c'_i}$, $(\forall M)[(\exists N)(\forall n>N)(\forall 1\leq m<M) \pi_n(m) = \kappa(m)$ and $(\exists N')(\forall n'>N')(\forall 1\leq m<M) \pi'_{n'}(m) = \kappa'(m)]$ thus $(\forall p>P=\max\{N,N'\})(\forall 1\leq m<M)$ \begin{align*} \kappa'\circ\kappa(m)&= \pi'_p\circ\pi_p(m)\\ &= [\circ_{i=1}^{P}(\sigma_{b'_i})^{c'_i}] \circ [\circ_{i=1}^{P}(\sigma_{b_i})^{c_i}](m)\\ &=\circ_{i=1}^{P}[(\sigma_{b'_i})^{c'_i} \circ (\sigma_{b_i})^{c_i}] \text{ (by commutativity of $\sigma_l$)} \end{align*} From this, we can immediately conclude $\kappa^{-1} = \lim_{n\to\infty}\circ_{i=1}^{n}(\sigma_{b_i})^{-c_i}$.

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Now to construct the isomorphism, Let $F:\mathbb{R}/\mathbb{Z}\longrightarrow A/B$, $$F(\theta+\mathbb{Z}) = F\left(\sum_{i=1}^{\infty}\frac{c_i}{b_i}+\mathbb{Z}\right) = [\circ_{i=1}^{\infty}(\sigma_{b_i})^{c_i}]\circ B$$ for any $c:\mathbb{N}\longrightarrow \mathbb{Z}$ and $b:\mathbb{N}\longrightarrow \mathbb{N}$ such that $\sum_{i=1}^{\infty}\frac{c_i}{b_i}\to \theta$. Now to show that $F$ is an isomorphism. Since the set of all convergent sequences if the form above converge to elements in $\mathbb{R}$, $F$ is trivially surjective. Suppose \begin{align*} F(\theta+\mathbb{Z}) &= F(\theta'+\mathbb{Z})\\ F\left(\sum_{i=1}^{\infty}\frac{c_i}{b_i}+\mathbb{Z}\right) &= F\left(\sum_{i=1}^{\infty}\frac{c'_i}{b'_i}+\mathbb{Z}\right)\\ [\circ_{i=1}^{\infty}(\sigma_{b_i})^{c_i}]\circ B &= [\circ_{i=1}^{\infty}(\sigma_{b'_i})^{'c_i}]\circ B\\ [\circ_{i=1}^{\infty}(\sigma_{b'_i})^{c'_i}]^{-1}\circ[\circ_{i=1}^{\infty}(\sigma_{b_i})^{c_i}]\circ B &= B\\ \circ_{i=1}^{\infty}[(\sigma_{b'_i})^{-c'_i}\circ(\sigma_{b_i})^{c_i}]\circ B &= B \\ \circ_{i=1}^{\infty}[(\sigma_{b'_i})^{-c'_i}\circ(\sigma_{b_i})^{c_i}] &\in B\\ \sum_{i}^{\infty}\frac{c_i}{b_i}-\frac{c'_i}{b'_i} &\equiv_1 0\\ \theta +Z &= \theta'+Z \end{align*} So $F$ is injective; additionally, $F$ is clearly a homomorphism. So now I am wondering where the mistakes are in my attempted proof? I'm sure there are several trivial mistakes, but I will edit those out as I find them. Can anyone find any nontrivial mistakes in my reasoning (or trivial mistakes/errors in notation)? Also I am beginning to understand Qiaochu Yuan's argument, so I will eventually add another edit to discuss how it relates to my argument.

Answer 1

I believe it is impossible to write down an embedding $S^1 \to \text{Aut}(\mathbb{N})$ in ZF so if your construction doesn't use some form of the axiom of choice I don't think it can work. You are working with the torsion subgroup of $S^1$, which is isomorphic to $\mathbb{Q}/\mathbb{Z}$. This subgroup is countable so it's easy to write down an explicit embedding of it into $\text{Aut}(\mathbb{N})$: it acts on itself, and you can write down an explicit bijection $\mathbb{Q}/\mathbb{Z} \cong \mathbb{N}$.

But $S^1$ is much larger (uncountably larger) than its torsion subgroup, and this action is very far from continuous in any reasonable sense, so I don't think there's any hope of "completing" it and still getting an action on a countable set. Ask me questions about my construction if you don't understand it!

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Edit (this has also been edited into my answer to the original question):

Okay, as suspected the answer to this question is independent of ZF. There's a model of ZF constructed by Shelah in which every set of real numbers has the Baire property. This implies, if I understand correctly, that there are no nonzero homomorphisms from $\mathbb{R}$ to any countable abelian group (since any countable abelian group with the discrete topology is a Polish group, so in this model any homomorphism from $\mathbb{R}$ to such a group is automatically measurable and so automatically continuous). So $\mathbb{R}$, and $SO(2)$, have no subgroups of countable index in this model.