The first thing to notice is that infinite permutations may have infinite support, that is, they may move infinitely many elements. Therefore, we cannot expect to express them as finite compositions of permutations having only finite support.

But if we allow (well-defined) infinite compositions, then the answer is that every permutation can be expressed as a composition of disjoint cycles and also expressed as a composition of transpositions. So the answer to question 1 is yes, and the answer to question 2 is no.

To see this, suppose that f is a permutation of ω. First, we may divide f into its disjoint orbits, where the orbit of n is defined as all the numbers of the form f^{k}(n) for any integer k. The action of f on each of these orbits commute with each other, because the orbits are disjoint. And the action of f on each such orbit is a cycle (possibly infinite). So f can be represented as a product of disjoint cycles. For the transposition representation, it suffices to represent each such orbit as a suitable product of transpositions. The finite orbits are just finite cycles, which can be expressed as a product of transpositions in the usual way. An infinite orbit looks exactly like a copy of the integers, with the shift map. This can be represented in cycle notation as (... -2 -1 0 1 2 ...). This permutation is equal to the following product of transpositions:

- (... -2 -1 0 1 2 ...) = [(0 -1)(0 -2)(0 -3)...][...(0 3)(0 2)(0 1)]

I claim that every natural number is moved by at most two of these transpositions, and that the resulting product is well-defined. On the right hand side of the equality, I have two infinite products of transpositions. Using the usual order of product of permutations, the right-most factor is first to be applied. Thus, we see that 0 gets sent to 1, and subsequently fixed by all later transpositions. So the product sends 0 to 1. Similarly, 1 gets sent to 0 and then to 2, and then unchanged. Similarly, it is easy to see that every non-negative integer n is sent to 0 and then to n+1 as desired. Now, the right-hand factor fixes all negative integers, which then pass to the left factor, and it is easy to see that again -n is sent to 0 and then to -n+1, as desired. So altogether, this product is operating correctly. An isomorphic version of this idea can be used to represent the action of any infinite orbit, and so every permutation is a suitable well-defined product of transpositions, as desired.

Thus, the answers to the questions in (1) are yes, and the answer to question (2) is no.

aredense in $\operatorname{Sym}(X)$. Because I was meeting a friend for dinner, I didn't want to be a few minutes late, so the comment I posted was not very insightful. It happens... $\endgroup$ – Pete L. Clark Mar 10 '10 at 8:53