symmetry in natural numbers

Question

Given a finite set $E$ we can associate a group of symmetries or permutation $S_n$ where $n=|E|$ is the cardinal of $E$. My question is what if the set is infinite or more precisely countable ? Is there any symmetries in the set of natural numbers $\mathbb{N}$ and by extension in countable sets ?

Answer 1

The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $\mathbb{N}$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: \mathbb{N}^* \longrightarrow \mathbb{N}^*$ defined by: $$f(n)=\begin{cases} n+1 & \mbox{if $n$ is odd} \\ n-1 & \mbox{if $n$ is even} \end{cases}$$ To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows: $$f_2(n)=\begin{cases} n+2 & \mbox{if} & n\equiv 1 \mod 3 \\ n & \mbox{if} & n\equiv 2 \mod 3 \\ n-1 & \mbox{if} & n\equiv 0 \mod 3 \end{cases}$$ And, $$f_3(n)=\begin{cases} n+3 & \mbox{if} & n\equiv 1 \mod 4 \\ n+1 & \mbox{if} & n\equiv 2 \mod 4 \\ n-1 & \mbox{if} & n\equiv 3 \mod 4 \\ n-3 & \mbox{if} & n\equiv 0 \mod 4 \end{cases}$$ One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : \mathbb{N}^* \longrightarrow \mathbb{N}^*$ where the pattern is now clear. $$f_k(n)=\begin{cases}\begin{cases} n+k-2i & \mbox{if} & n\equiv i+1 \mod (k+1) & \mbox{for} & i=0,1,\ldots,\frac{k}{2} \\ n-k+2i & \mbox{if} & n\equiv i+1 \mod (k+1) & \mbox{for} & i=\frac{k}{2}+1,\ldots,k \end{cases} & \mbox{if $k$ is even} \\ \begin{cases} n+k-2i & \mbox{if} & n\equiv i+1 \mod (k+1) & \mbox{for} & i=0,1,\ldots,\frac{k-1}{2} \\ n-k+2i & \mbox{if} & n\equiv i+1 \mod (k+1) & \mbox{for} & i=\frac{k-1}{2}+1,\ldots,k \end{cases} & \mbox{if $k$ is odd} \end{cases}$$ One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $\mathbb{N}^*$ was partitioned into a union of disjoints sets. Namely $\mathbb{N}^*=\cup_{m \in \mathbb{N}^*}E_{m,k}$, where $E_{m,k}=\{m(k+1)-k,m(k+1)-k+1,\ldots,m(k+1)\}$ with the same cardinality of each set, $|E_{m,k}|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_{m,k}$ noted ${f_{k}}_{|E_{m,k}}$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation $${f_4}_{|E_{2,4}}= \begin{pmatrix} 6 & 7 & 8 & 9 & 10 \\ 10 & 9 & 8 & 7 & 6 \end{pmatrix} \qquad {f_5}_{|E_{3,5}}= \begin{pmatrix} 13 & 14 & 15 & 16 & 17 & 18 \\ 18 & 17 & 16 & 15 & 14 & 13 \end{pmatrix}$$ The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.

Definition (Permutation in countable set): Let $E$ be a countable set, and $(E_n)_{n=1,\ldots , \infty}$ is a disjoint partition of E composed by finite sets that is $$ E=\cup_{n \in \mathbb{N}}E_n \qquad \mbox{,} \qquad \cap_{n \in \mathbb{N}}E_n = \emptyset \qquad \mbox{and} \qquad |E_n| < \infty$$ A bijection $\sigma : E \longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ \sigma_{|E_n}(E_n)=E_n \qquad \mbox{and} \qquad \sigma_{|E_n} \in S_{|E_n|}$$ where $S_{|E_n|}$ is the group of permutations of order $|E_n|$.

Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $\sigma$ as defined which we denote $\Sigma_p(E)$ with the composition operation is a group.

Thank you.