Results 1 to 9 of 9

Math Help - Show that S4 has a unique subgroup of order 12.

  1. #1
    nhk
    nhk is offline
    Junior Member
    Joined
    Apr 2010
    Posts
    43

    Show that S4 has a unique subgroup of order 12.

    I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by nhk View Post
    I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?
    first show that A_4 has no subgroup of order 6. then, since |HA_4|=\frac{144}{|H \cap A_4|} \mid 24 and |H \cap A_4| < 12, we must have |H \cap A_4|=6. contradiction!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by nhk View Post
    I know that A4 is a subgroup of order 12. So I tried to prove that S4 has a unique subgroup of order 12 by assuming that there existed another subgroup of order 12, lets call this other subgroup H. The hint I was given was to also assume that the subgroup H was normal. Any tips on where I should go from here?
    I don't know how much you done have on group theory already, but the subgroup you are dealing with has index 2, and is denoted A_4. It is called the alternating group of degree 4. the question asks you to assume the subgroup is normal because it is; every subgroup of index 2 in a group is normal (essentially, because there can only exist two cosets, and so left and right cosets must be equal).

    Now, permutations in S_n are special. A permutation of a specific cycle type* is conjugate to every other permutation of that cycle type. (g is conjugate to h if there exists some element a such that g = a^{-1}ha). For example, (12)(123)(12) = (132). This should hopefully be in your notes. If it is not, then say and someone will have to come up with a different method to solve this probelm!

    A normal subgroup is a group which is closed under conjugation. So in S_n, if there exists a permutation of a cycle type then all permutations of this cycle type are in the normal subgroup.

    Let us now count the number of each cycle types,

    There is 1 cycle of type () in S_4.

    There are 4.3/2 = 6 cycles of type (12) in S_4.

    There are 4.3.2/3 = 8 cycles of type (123) in S_4.

    There are 4.3.2.1/4 = 6 cycles of type (1234) in S_4.

    There are 4.3.2.1/(2.2.2) = 3 cycles of type (12)(34) in S_4.

    (for a normal cycle you divide by the length as (12) = (21), (123) = (231) = (312) etc. For the last part you divide by 2.2.2 as two of the twos are because (12)=(21) and the third is because (12)(34) = (34)(12). Alternatively, just sum up all the other cycle types and subtract from 24, 24-(6+8+6+1) = 3.)

    Note that these all sum up to 24. Now, we wish to sum together the numbers 1, 6, 8, 6 and 3 to get 12 in two different way. However, there is only one way of doing this. This means there can only be one subgroup of order 12 in S_4.

    *Note that by cycle type I mean (123)(45), (123), (12)(34 are all different while (123) and (132) have the same cycle type.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    first show that A_4 has no subgroup of order 6. then, since |HA_4|=\frac{144}{|H \cap A_4|} \mid 24 and |H \cap A_4| < 12, we must have |H \cap A_4|=6. contradiction!
    Do you mean normal subgroup? As there is an obvious copy of A_3 in there...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Swlabr View Post
    Do you mean normal subgroup? As there is an obvious copy of A_3 in there...
    the order of A_3 is 3 not 6.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    the order of A_3 is 3 not 6.
    Good point, not sure why I read it as S_4.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    nhk
    nhk is offline
    Junior Member
    Joined
    Apr 2010
    Posts
    43
    Quote Originally Posted by NonCommAlg View Post
    first show that A_4 has no subgroup of order 6. then, since |HA_4|=\frac{144}{|H \cap A_4|} \mid 24 and |H \cap A_4| < 12, we must have |H \cap A_4|=6. contradiction!
    sorry for taking so long to respond, but why is it that we can use those set of equations that you have up there, is it because both of the subgroups of order 12 are normal?
    Thanks again for your help, it has been greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by nhk View Post
    sorry for taking so long to respond, but why is it that we can use those set of equations that you have up there, is it because both of the subgroups of order 12 are normal?
    Thanks again for your help, it has been greatly appreciated.
    Do you mean the |AB| = \frac{|A|\cdot|B|}{|A \cap B|}? This is a well-known formula and holds if A, B and AB are all subgroups.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    nhk
    nhk is offline
    Junior Member
    Joined
    Apr 2010
    Posts
    43
    Quote Originally Posted by Swlabr View Post
    Do you mean the |AB| = \frac{|A|\cdot|B|}{|A \cap B|}? This is a well-known formula and holds if A, B and AB are all subgroups.
    THanks, i thought that formula was valid but I could'nt find it in my book. Must have been looking in the wrong section. Thanks for your help and the fast response
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 2nd 2011, 09:07 PM
  2. Show that cosx = lnx has a unique solution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 30th 2010, 06:17 AM
  3. Order of a Group with a Unique Subgroup
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 18th 2010, 08:53 PM
  4. Prove that a group of order 375 has a subgroup of order 15
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 14th 2010, 12:08 AM
  5. Show that f(x)=g(x) has a unique solution
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 18th 2010, 02:44 PM

Search Tags


/mathhelpforum @mathhelpforum