Defining parity on the set of all bijections of ${\mathbb{N}}$

Question

Let $\phi:{\mathbb{N}}\rightarrow {\mathbb{N}}$ be a bijection. Can we extend the notion of parity (of a finite permutation) to $\phi$ ?

In other words, Can we define a group homomorphism $\Lambda $, between the group of all bijections of $\mathbb{N}$ to the group {1,-1,$\cdot$} such that $\Lambda(\psi)=-1$ if $\psi$ is a "transposition":

There exists $n_{1},n_{2}\in \mathbb{N}$, where $n_{1}\neq n_2$ and $\psi(n_{1})=n_{2} ,\psi(n_{2})=n_{1}$ and ${\psi(x)=x , \forall x \neq n_{1},n_{2} }$.

My guess is that building such a homomorphism requires some version of the axiom of choice / ultra-filters.

Answer 1

It would be easier, and equivalent, to consider $\mathbb{Z}.$ But I'll stick to $\mathbb{N}.$

Consider $\sigma=(2\ 3)(4\ 5)(6\ 7)\cdots$ (so $0$ and $1$ are fixed points) and $(0\ 1)\sigma.$ They have the same cycle type but should not have the same parity. Note that fixed points and the particular ordering of $\mathbb{N}$ should be irrelevant.

For a similar example with conjugation consider $\tau=(5\ 7)(9\ 11)(13\ 15)\cdots$ and $\tau'=(1\ 3)(5\ 7)(9\ 11)(13\ 15)\cdots$ which should again not gave the same parity. To turn $\tau$ into $\tau',$ conjugate by this product of two infinite cycles $$(\cdots 13\ 9\ 5\ 1\ 0\ 4\ 8 \cdots) (\cdots 11\ 7\ 3\ 2\ 6\ 20 \cdots).$$

Answer 2

The signature homomorphism, defined on finitely supported permutations, does not extend to the group of all permutations.

To show this, it's enough to write a transposition as a commutator. Namely, call X-element a permutation consisting of infinitely many cycles of each finite length (and no infinite cycle). Then it's quite clear that the product of any X-cycle $c$ with any transposition $t$ (or any finitely supported element) is also a X-cycle, and that all X-cycles are conjugate. Thus one can write $tc=bcb^{-1}$, so $t=[b,c]$.