Can we find an element of infinite order in a symmetric group of infinite order?

Question

In particular, I'm thinking of a simple example: the group $S_\Omega$ given $\Omega = \{1, 2, 3, ...\}$.

I've been thinking of elements of $S_\Omega$ in terms of their cycle decomposition, which may be the problem. But, in particular, I can't seem to think of a bijection $\varphi : \mathbb{N} \to \mathbb{N}$ that permutes the elements of $\mathbb{N}$ infinitely. For example, I can't define $\varphi (x) = x+1$ for all $x \in \mathbb{N}$ because there does not exist a $y \in \mathbb{N}$ such that $\varphi (y) = 1$.

Would the situtation be different if $\Omega$ was the set of integers and thus infinite in both the positive and negative directions?

Answer 1

Consider the permutation $$(1)(2\;3)(4\;5\;6)(7\;8\;9\;10)\cdots$$

Answer 2

For $\mathbb Z$ the function $\sigma(x)=x+1$ works.

Now pick $f : \Omega \to \mathbb Z$ a bijection. Then $f^{-1} \circ \sigma \circ f \in S_\Omega$ has the desired properties.

Answer 3

$f(x) = x - 2$ if $x \ge 4$ is even, $f(2) = 1$, $f(x) = x + 2$ if $x$ is odd.