1
$\begingroup$

The permutation group $S_n$ can be generated by an element $$(i \; \; i+1)$$ together with an $n$-cycle.

First, I would like to check that the above statement is indeed true?

My main question is, is it known what the relations are when we generate $S_n$ like this?

$\endgroup$
1
4
$\begingroup$

Well, let's say that first one is on you. For the other one, one can (somewhat easily) find out that $$ \langle t, c \,|\,t^2, c^n, (tc)^{n-1}, [t, c]^3, [t, c^k]^2 = 1 \text{ for } 2 \leq k \leq n/2\rangle $$

is indeed the presentation you need. The best (to my knowledge) way to check its validity is first writing down Coxeter presentation as a reflection group, and then eliminating extra generators replacing them by products of conjugates of a transposition.

$\endgroup$
6
  • $\begingroup$ Where square brackets denotes the commutator? $\endgroup$ – Matt Jan 4 at 3:46
  • 2
    $\begingroup$ Yes. A forewarning for a person diving into group theory literature: always note if the author right or left, i. e. uses $[a, b] = a^{-1}b^{-1}ab$ or $[a, b] = aba^{-1}b^{-1}$. It does not matter in my answer, but reading two texts on same subject with different chirality can get quite annoying. :) $\endgroup$ – xsnl Jan 4 at 3:53
  • $\begingroup$ Thank you!!! :) $\endgroup$ – Matt Jan 4 at 11:06
  • $\begingroup$ I don't believe this is the correct presentation for all $n$ (doesn't work for $n=6$ for example). $\endgroup$ – Steve D Jan 4 at 19:02
  • 1
    $\begingroup$ Oh I see, you have a typo: should be $[t,c^k]^2$ in the last relation. $\endgroup$ – Steve D Jan 4 at 19:13

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.