15
$\begingroup$

This is a nice question which I don't have any clue.

Let $S_{\infty}$ be the group of all permutations of $\mathbb{N}.$

Is it possible to embed $SO(3)$ into $S_{\infty}?$

$\endgroup$
6
  • 7
    $\begingroup$ Found a paper that claims that the answer is yes: matwbn.icm.edu.pl/ksiazki/fm/fm164/fm16412.pdf $\endgroup$
    – Dan Shved
    Apr 18, 2013 at 7:39
  • $\begingroup$ Since $SO(3)$ is simple, it is sufficient to prove that it has a subgroup of countable index. No idea how to do that though. $\endgroup$
    – Dan Shved
    Apr 18, 2013 at 7:43
  • $\begingroup$ Dear @Dan Shved, that paper is great! you should write it as an answer. $\endgroup$ Apr 18, 2013 at 9:19
  • $\begingroup$ Am I missing something here...if $G$ is an infinite group then does $G$ not embed into $S_{|G|}$? Or do you have to go up a cardinal? $\endgroup$
    – user1729
    Apr 18, 2013 at 9:40
  • 1
    $\begingroup$ @user1729: No need to go up a cardinal, but $|SO(3)| = |\mathbb{R}|$, not $|\mathbb{N}|$. $\endgroup$
    – Martin
    Apr 18, 2013 at 9:43

1 Answer 1

4
$\begingroup$

The answer is yes, as proved in this paper:

http://matwbn.icm.edu.pl/ksiazki/fm/fm164/fm16412.pdf

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .