Given a set, $X$, let $H(X)$ denote the subgroup of $Sym(X)$ generated by the elements of order $2$.
- If $X$ is finite, $Sym(X)$ is generated by transpositions, so $H(X)=Sym(X)$.
- I think that $H(X)=Sym(X)$ when $X$ is uncountable, too. The argument essentially comes to down to showing that one can construct a permutation of each possible cycle type.
- If $X$ is countably infinite, I think $H(X)$ contains every element with the following property $(\star)$: When $\sigma$ is expressed as the product of (possibly infinitely many) disjoint cycles, none of these cycles is co-finite. I suspect that either $H(X)=Sym(X)$ or $H(X)=\{\sigma:\sigma\text{ satisfies }\star\}$, but I haven't found a way to determine which case it is. For example, I haven't determined whether $H(\mathbb{Z})$ contains the permutation $n\mapsto n+1$ (which does not satisfy $\star$).
So, my questions are: Does $H(X)=Sym(X)$ for all $X$? And, if not, what do the exceptions look like?
Best Answer
Every cycle is the product of two elements of order two.
In particular, let $\sigma$ and $\tau$ be the permutations $n \mapsto n+1$, and $n \mapsto -n$ of ${\mathbb Z}$, respectively.
Then (composing from left to right), $\sigma\tau$ is $n \mapsto -n-1$, and $\sigma = (\sigma\tau)\tau$, where $\sigma\tau$ and $\tau$ both have order $2$.