4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.