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Let $\sigma$ be a permutation of $\mathbf Q.$ We call $\sigma$ bounded (the term might be somewhat misleading, but however it is used in a couple of papers) if there is a real number $M$ such that $$ \mathrm{dist}(\sigma x,x) =|\sigma x-x| \le M $$ for all $x \in \mathbf Q.$

Now can a bounded infinite cycle $$ \pi=(\ldots,-2,-1,0,1,2,\ldots) $$ be written as a product of bounded involutions (permutations of order two)? Notice that "nice" bounded products of, say two, disjoint infinite cycles can be written as products of bounded involutions, e.g. the permutation $$ (\ldots,-4,-2,0,2,4,\ldots) (\ldots,-3,-1,1,3,\ldots). $$

EDIT: to clarify a bit chandok's argument below: suppose $\pi$ is a product of two involutions $\sigma$ and $\tau$: $\pi=\sigma \tau.$ Then $$ \sigma \tau \sigma(n+1)=\sigma(n+1)+1. $$ for all integers $n.$ On the other hand, $n+1=\pi(n)$ and $\tau \sigma=\pi^{-1},$ whence $$ \sigma (\tau \sigma) \pi(n) = \sigma \pi^{-1} \pi(n)=\sigma(n)=\sigma(n+1)+1. $$ Thus $\sigma(n+1)=\sigma(n)-1$ and $\sigma(n)=a-n$ for all naturals $n.$ Thus $\sigma$ is unbounded.

The argument seems to be not easily adaptable for products of more than two involutions.

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Suppose you have involutions $\sigma, \tau$ such that $\sigma \tau (n) = n+1$ forall $n \in \mathbb{Z}$, and $\sigma \tau (n) = n$ otherwise

Then forall $n \in \mathbb{Z}$, we have $\sigma \tau (n) = n+1$, so $\tau (n) = \sigma\sigma\tau(n) = \sigma(n+1)$, and $\sigma\tau\sigma(n+1) = \sigma\tau\tau(n) = \sigma(n)$. Moreover, $\sigma \tau (n-1) = n \neq n+1 = \sigma \tau (n)$, so $\sigma(n) = \tau(n-1) \neq \tau(n) = \sigma(n+1)$, so $\sigma(n+1)$ is not fixed by $\sigma \tau$.

Therefore $\sigma(n+1)$ is an integer, and $\sigma(n+1)+1 = \sigma(n)$, and likewise for $\tau$. So there exists integers $a$ and $b$ such that $\sigma(n) = a-n$ and $\tau(n) = b-n$. And these involutions are not bounded.

Edit : So it can't be written as a product of only two involutions (I missed the part not being restricted to products of more than 2 involutions, I will think about it)

As you noticed, using a product of 2 bounded involutions you can form the product of two infinite cycles going opposite directions. For $(a,b) \in \mathbb{R}$, call $\tau_{(a,b)} = \prod_{n \in \mathbb{Z}} ((n+a) (n+b))$. $\tau_{(a,b)}$ is a bounded involution, and $\sigma_{(a,b)} = \tau_{(a+1,b)} \tau_{(a,b)}$ is a permutation that adds $1$ to numbers of the form $n+a$ and substracts $1$ to numbers of the form $n+b$.

Now consider the (infinite) product $\sigma = \sigma_{(0,1/2)} \sigma_{(1/2,1/3)} \sigma_{(1/3,1/4)} \ldots$. $\sigma$ is just the permutation you want.

To write it as a finite product of bounded involutions, first rewrite it as a product of two permutations so that the things inside a product all have disjoint support so the involutions later can commute between themselves : $\sigma = (\sigma_{(0,1/2)} \sigma_{(1/3,1/4)} \ldots) (\sigma_{(1/2,1/3)} \sigma_{(1/4,1/5)} \ldots)$

Then expand each $\sigma_{(a,b)}$ into its product of two involutions $\sigma = ((\tau_{(1,1/2)}\tau_{(0,1/2)})(\tau_{(4/3,1/4)}\tau_{(1/3,1/4)})\ldots) ((\tau_{(3/2,1/3)}\tau_{(1/2,1/3)})(\tau_{(5/4,1/5)}\tau_{(1/4,1/5)}) \ldots)$

Commute them to get a product of 4 bounded involutions (with infinite support):

$\sigma = (\tau_{(1,1/2)}\tau_{(4/3,1/4)}\ldots)(\tau_{(0,1/2)}\tau_{(1/3,1/4)}\ldots) (\tau_{(3/2,1/3)}\tau_{(5/4,1/5)}\ldots)(\tau_{(1/2,1/3)}\tau_{(1/4,1/5)}\ldots)$

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This does not answer the OP's question. It might be that $\sigma\tau$ can be written as some other product of involutions that are bounded. It is clear that any permutation can be written as a product of involutions some of which are unbounded. Your example is an instance of that. – Alex B. Apr 4 '11 at 17:29
thanks! @Alex: Right. Moreover, what I asked was about productS, products of two, three, four, etc. bounded involutions. – Olod Apr 4 '11 at 17:38
A neat idea! Thank you very much indeed. – Olod Apr 4 '11 at 20:40

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