Results 1 to 3 of 3

Math Help - Smallest Subgroup of Sn

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    13

    Question Smallest Subgroup of Sn

    Good afternoon fellow math goers.
    I entered this problem a week ago and with helpful hints have made what I think is progression on this rather challenging problem. I need some advice on my proof though so any more help would be greatly appreciated.


    Problem: Prove that the smallest subgroup of Sn containing (1, 2) and (1, 2, . . . , n) is Sn.

    Conjunction:
    (12...n)(12)(12...n)^-1=(23)
    (12...n)(23)(12...n)^1=(34)
    (12...n)(34)(12...n)^-1=(45)
    WLOG (12...n)(n-2 n) (12...n)^-1=(n-1 n)
    (12...n)^n-1 (n-1 n)(12...n)^n-1=(n1)

    So essentially this gives us a difference k=0 and that we proved its a generator of Sn correct? Now should I use induction and do k+1 terms or should I do the following:

    Suppose you have some group such that ,
    Above I proved that (12)(1...n) generates Sn so therefore

    How do I prove that
    . so I can establish that G=Sn??

    AHHHH






    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by RoboMyster5 View Post
    Good afternoon fellow math goers.
    I entered this problem a week ago and with helpful hints have made what I think is progression on this rather challenging problem. I need some advice on my proof though so any more help would be greatly appreciated.

    Problem: Prove that the smallest subgroup of Sn containing (1, 2) and (1, 2, . . . , n) is Sn.

    Conjunction:
    (12...n)(12)(12...n)^-1=(23)
    (12...n)(23)(12...n)^1=(34)
    (12...n)(34)(12...n)^-1=(45)
    WLOG (12...n)(n-2 n) (12...n)^-1=(n-1 n)
    (12...n)^n-1 (n-1 n)(12...n)^n-1=(n1)

    So essentially this gives us a difference k=0 and that we proved its a generator of Sn correct? Now should I use induction and do k+1 terms or should I do the following:

    Suppose you have some group such that ,
    Above I proved that (12)(1...n) generates Sn so therefore
    How do I prove that . so I can establish that G=Sn??

    AHHHH




    From my understanding, you used the theorem that S_n is generated by n-1 transpositions (1,2), (2,3), ..., (n-1, n). You showed that (1,2) and (1,2, ... , n) indeed generate those n-1 transpostions, which shows that S_n \subseteq G. We see that G (a set generated by (1,2) and (1,2,...,n) ) is just the set of permutations of {1,2, ..., n}. By the definition of the symmetric group of degree n, it forces G \subseteq S_n. Thus G=S_n, a symmetric group of degree n.

    EDIT (some additional remarks): G must contain (1,2) and (1,2, ... , n). Since G is the group by hypothesis, G must contain a set generated by (1,2) and (1,2, ... ,n). Since our choice of G should be minimal, if we show that a set generated by (1,2) and (1,2, ... ,n) is the group, the proof is completed.
    Last edited by aliceinwonderland; December 9th 2009 at 01:42 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Wow I think you just made this make complete sense...I applaud you. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 13
    Last Post: May 12th 2011, 04:32 AM
  2. Replies: 2
    Last Post: March 2nd 2011, 08:07 PM
  3. characterisitic subgroup implies normal subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 8th 2010, 03:13 PM
  4. Replies: 1
    Last Post: March 17th 2010, 06:10 PM
  5. Replies: 2
    Last Post: March 26th 2009, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum