Let $X$ be a set, potentially infinite, and $\tau$ a transposition of $X$, meaning \begin{align*} \tau(x)&=y \\ \tau(y)&=x \\ \tau(z)&=z \quad \text{for all} \quad z\notin\{x,y\} \end{align*} for some distinct $x,y\in X$

Show that there are no bijections $f,g:X \rightarrow X$ satisfying $\tau=f\circ g \circ f^{-1} \circ g^{-1}$.

I stumbled on this while looking for alternative definitions of parity of (finite) permutations, I don't know how to prove it, and my claim could be false.

A similar question (one I can solve) would be to show that $\tau$ is not a square but I believe the above problem requires a more sophisticated argument.

My idea was that to define parity the abelianization morphism. Showing that a product of two transpositions is in the kernel is pretty standard, the above would be a step to proving the kernel is not all $S_n$.

Of course this definition is rather convoluted. If you have any unusual definition for the parity of permutations of $X$, specially one that never involves ordering the elements of $X$, I would be very interested.

  • $\begingroup$ As Henning Makholm poijnted out, this is false. Note however that $\tau$ is not a commutator of two bijections $f,g$ of finite support. The restricted symmetric group on $X$ consists of all permutations of $X$ with finite support. As is the case for finite $X$, this has a normal subgroup of index $2$ consisting of even permutations, which is simple when $|X| \ge 5$. $\endgroup$ – Derek Holt Oct 6 '17 at 13:23

It is not true. As a counterexample, consider

  • $X=\mathbb Z$
  • $f(n)=n+2$
  • $g$ swaps $2n$ with $2n+1$ for $n\ge 0$ and leaves negative numbers unchanged.

In fact the group of permutations of an infinite set has the strong property that every element is a commutator, though this takes some ingenuity to prove.

  • 2
    $\begingroup$ Thank you for preventing me from making a fool of myself. I was about to post a proof that it is true, when I saw your answer. $\endgroup$ – José Carlos Santos Oct 6 '17 at 11:16

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