Let $X$ be an infinite set. Let $\operatorname {Sym}(X)$ denote the group of all bijections from $X$ onto itself. I have been thinking about the existence of elements of infinite order in this group.
For $\operatorname{card}(X)=\aleph_0,$ I've found the following construction. Let us identify $X$ and $\mathbb{N}.$ Let
$ \operatorname{Sym}(\mathbb N)\ni\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & ...\\ 2 & 1 & 4 & 5 & 3 & 7 & 8 & 9 & 6 & 11 & ... \end{pmatrix}. $
It is clear that for any natural number $n$ there exists another natural number $k$ such that
$ \alpha^n(k)\neq k. $
This is an example of an element of infinite order in $\operatorname{Sym}(\mathbb N).$ I was thinking if I could find such an element for any infinite set $X.$ I found this. Let $\leq$ be a well-ordering of $X.$ Let $x_0$ be the least element of $X.$ Let $x+n$ denote the successor of $x+(n-1)$ if such exists, and $x_0+0=x_0.$ Let $Y=\{x_0+n\,|\,n\in \mathbb{N}\cup\{0\}\}.$
This is well-defined because $X$ is an infinite well-ordered set. We define $\beta\in\operatorname{Sym}(X)$ as follows.
$ \begin{array} & \beta(x_0+n)=x_0+\alpha(n+1)-1 & \mbox{for } n\in \mathbb N\cup \{0\}\\ \beta(x)=x & \mbox {for } x\not\in Y \end{array} $
This, if I'm not mistaken, is an example of an element of infinite order in $\operatorname{Sym}(X).$ However, it doesn't meet my needs. I would like to find an example with few or no fixed points. I'm having trouble with formalizing my reasoning here.
I would like to partition the set $X$ into copies of $\mathbb{N}$ and essentially use the first example on every one of them. To my understanding, any well-ordered set consists of, possibly unimaginably many, copies of $\mathbb N$ put one after the other, then copies of the result of the previous operation put one after the other, and so on, with possible tails of smaller "order" (or whatever it is called) at the end. Even if something is wrong with my understanding, I'm quite sure the partition I'd like to have exists. But I'm having a lot of trouble writing down a formal proof of this because I have never done anything with ordinals at all.
So my first question is if you could help me formalize this reasoning. I'm not asking for a very formal proof of course. Just something that would convince me, because what I have written does not.
My second question is if this can be done without the use of the axiom of choice. More precisely:
1) Is it possible to prove without assuming the axiom of choice that for any infinite set $X,$ there is an element $\alpha\in\operatorname{Sym}(X)$ of infinite order?
2) Is it possible to prove without assuming the axiom of choice that for any infinite set $X,$ there is an element $\alpha\in\operatorname{Sym}(X)$ of infinite order, such that the set of fixed points of $\alpha$ has cardinality lesser than $\operatorname{card}(X)?$
I would be very grateful if the answerers would kindly use simple language if it's possible. I'm very far from knowing much about these things.