Originally Posted by

**tonio** Good question: I think it does, but a combinatorial proof isn't available unless we limit ourselves to permutations

moving a finite number of objects (because in that case we can work within $\displaystyle S_n$ , for some natural n).

So assume $\displaystyle \pi\in S_X\,,\,\,|X|\geq \aleph_0\,,\,\,\pi(x)\neq x\,\,\forall x\in S $ , for at least an infinite countable set of elements $\displaystyle S\subset X$.

But then we can find say $\displaystyle x_1,x_2,x_3\in S$ s.t. $\displaystyle \pi(x_i)\neq x_i\Longrightarrow \exists \sigma\in S_{X'_3}\,,\,\,X'_3:=\{x_i\,,\,\pi(x_i)\;;\;i=1,2, 3\}\,,\,s.t\,\,\sigma\overline{\pi}\neq \overline{\pi}\sigma$ , with $\displaystyle \overline{\pi}:=\pi\mid_{S_{X'_3}}$.

As we can take $\displaystyle \sigma$ as element of $\displaystyle S_x$ we're then done.

Tonio