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

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?

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Do you know that $\Bbb N$ can be brought into bijection with $\Bbb Z$? For $\Bbb Z$ I think you can see a permutation with infinite order right away. – Marc van Leeuwen Apr 13 '14 at 7:47

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

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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.

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Interesting. Does this sort of conjugation work in general for bijections assuming that the bijections work for a set that contains the one in question? – Danny Apr 13 '14 at 6:15
@Danny If $A,B$ are any sets and $f: A \to B$ is a bijection, then $\sigma \in S_A$ and $f \circ \sigma \circ f^{-1}$ have the same order. – N. S. Apr 13 '14 at 6:19
Makes sense, thank you! – Danny Apr 13 '14 at 6:25

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

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