Show that if $A=\{1, 2,...\}$, then $S_A$ is an infinite group

Question

I want to show that if $A=\{1, 2,...\}$, then $S_A$ is an infinite group.

My proposed solution:

We consider the element $x=(1, 2,..., i, i+1,...)$ of $S_A$ , where we have used cyclic notation for representing elements of permutation groups, noting that there exists no positive finite integer $n$, such that $|x|=n$, we conclude that the cyclic subgroup of $S_A$ generated by $x$ is an infinite group, thus we conclude that $S_A$ is an infinite group.

Is my solution reasonable? (This is not a Homework problem, that I want to get solved unfairly )

Answer 1

No.

Your $x$ is not a bijection from $A$ to $A$. What maps to $1$? Thus $x\notin S_A$.

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There is infinitely many distinct elements of the form $(i,i+1)$ in $S_A$ for $i\in A$.

Answer 2

You can have an infinite cycle, but it needs to be infinite in both directions, as in @Derek Holt's comment. Otherwise not all points are "hit".

Alternative argument:

For any $n\in \Bbb N,$ we have $S_n\le S_\Bbb N\implies \lvert S_\Bbb N\rvert \ge n!.$