The Group of Bounded Permutations

Question

Let $$ S_B = \{\phi\in S_\mathbb{N} \mid \exists R\in\mathbb{N}: |\phi_n-n|\leq R\}. $$

Does anyone know if this subgroup has been well studied? Has it been given a name in the literature? I am hoping for a full characterization of its normal subgroups. The only ones I have found are those inherited by $S_\mathbb{N}$ (those in the Schreier-Ulam theorem).

Answer 1

Here is a construction of many normal subgroups of $S_B$. Given two increasing functions $f, g : \mathbb N \to [0, \infty)$, let us write $f \prec g$ if there are constants $a, b$ such that $f(x) \le a g(x + b)$ for all $x \in \mathbb N$. Now given an increasing function $f : \mathbb N \to [1,\infty)$ let$\def\supp{\operatorname{supp}}$

$$N_f = \{\pi \in S_B \mid |\supp(\pi) \cap [0,n]| \prec f\}.$$

Since $\supp(\pi\sigma) \subset \supp(\pi) \cup \supp(\sigma)$ it is clear that $N_f$ is a subgroup and if $\phi \in S_B$ and $|\phi(n) - n| \le R$ for all $n$ then $|\supp(\pi^\phi) \cap [0,n]| \le |\supp(\pi) \cap [0, n+R]|$, so $N_f$ is actually a normal subgroup. Moreover $N_f$ contains the subgroup $\mathrm{FSym}(\mathbb N)$ of all finitely supported permutations.

By considering $f(n) = n^\alpha$ for $0 < \alpha < 1$, we get continuum-many normal subgroups in this way (note that $n^\alpha \prec n^\beta$ iff $\alpha \le \beta$). With other choices of $f$ we can also get a large collection of pairwise incomparable normal subgroups (neither contained in the other).

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Edit: Here is a contruction of $2^c$ pairwise incomparable normal subgroups of $S_B$ (where $c = 2^{\aleph_0}$). Let $\beta \mathbb N$ be the set of ultrafilters. It is well-known that $|\beta \mathbb N| = 2^c$ (for example this follows from the fact that $[0,1]^{[0,1]}$ is separable). Now for any $p \in \beta \mathbb N$ consider

$$N_p = \{\pi \in S_B \mid \limsup_{n \to p} |\supp(\pi) \cap [0,2^{2^n}]| / 2^{2^n} = 0\}.$$

It follows exactly as above that $N_p \trianglelefteq S_B$ for all $p \in \beta\mathbb N \setminus \mathbb N$. Now consider two ultrafilters $p, q \in \beta\mathbb N \setminus \mathbb N$. Let $X \in p \setminus q$. Let $\pi \in S_B$ the product of all transpositions $(2i, 2i+1)$ where $2^{2^{n-1}} \le 2i < 2^{2^n}$ for $n \in X$. Then $\pi \in N_q \setminus N_p$, which demonstrates that $N_q \not\le N_p$.

Answer 2

Thanks to the answer from @Sean Eberhard, I found this normal proper subgroup, which is the largest I could find. $$N_O=\{\phi\in S_B\mid n\neq O\left(|\mathrm{supp}(\phi)\cap[0,n]|\right)\},$$ where $O$ denotes the big O operator.