I am illiterate in algebra/set theory, so I hope you will forgive my ignorance.

It is known that the symmetric group of a finite set is generated by elementary permutations. More precisely, $Sym\{1,...,n\}$ is generated by $(1,2),...,(n-1,n)$. Now to what extend does this generalize to infinite sets?

a. For instance, is $Sym(\mathbb{N})$ generated by $(1,2),...$ ? I know that the group is of cardinality at least continuum (take Riemann's theorem on conditionally converging sequences). But so is that of the set of natural sequences, so I have some hope for an injective map. If not, what would be a good generalization of the construction that works?

b. What about $Sym(\mathbb{R})$? I am not even sure what 'generate' would mean in this case, a composition of a continuum number of elements? Say, a function $\mathbb{R}\mapsto G$, where G is the set of generators. Can $Sym(\mathbb{R})$ be injectively mapped into the set of maps $\mathbb{R}\mapsto G$ for some meaningful $G\subset Sym(\mathbb{R})$? Can $G$ be

$$ G=\{P_{x,y}| x\in\mathbb{R}, y\in\mathbb{R}_+\},\quad P_{x,y}(t)=x\chi_{\{x-y\}}(t)+(x-y)\chi_{\{x\}}(t)+t(1-\chi_{\{x-y\}}(t)-\chi_{\{x\}}(t)), $$

where $\chi_A$ is the characteristic function of $A\subset\mathbb{R}$. In other words, $P_{x,y}$ just permutes $x$ and $x-y$.

If not, what would be a working alternative? Thank you.

  • 2
    $\begingroup$ For question a, the thing is that if the elementary permutations generate the group, then every element would be a finite product of those, so you should be looking at finite sequences (or sequences the are non-zero only for a finite number of terms), and this is countable, so no hope here. $\endgroup$ – Captain Lama Mar 23 '16 at 22:43
  • $\begingroup$ Thanks for the comment. I agree that the standard definition of 'generate' means 'in terms of finite products', in which case it does not work. But here I am trying to find a generalization that works. Some sort of infinite products... $\endgroup$ – Bedovlat Mar 23 '16 at 22:48
  • 1
    $\begingroup$ @Bedovlat: You'd have to take some care in defining such infinite products. For instance, $(1,2)(2,3)(3,4)\ldots$ doesn't correspond to a function on $\mathbb N$, since $1$ isn't mapped to any finite number. $\endgroup$ – joriki Mar 23 '16 at 22:59
  • $\begingroup$ A good remark, actually. Thanks. Probably as usual one needs certain topology to define the convergence of infinite products, in which case that sort of products won't converge. For instance, one could count only those products where the image of every natural number eventually stabilizes. $\endgroup$ – Bedovlat Mar 23 '16 at 23:04
  • 1
    $\begingroup$ @Bedovlat: That raises the question what it means to "count only certain products". This seems to be going in the direction that you don't have (as one does in group theory) a prior notion of "generate" and check whether a certain set is generated, but conversely would have to define the allowed products by the fact that they result in a well-defined bijection on $\mathbb N$. In that case, $\operatorname{Sym}(\mathbb N)$ is indeed "generated" by transpositions (and thus by adjacent transpositions), since you can go through the natural numbers in order, move them in place and leave them there. $\endgroup$ – joriki Mar 23 '16 at 23:15

Your Answer

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

Browse other questions tagged or ask your own question.