1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal groups

  1. Nov 8, 2008 #1

    I want to determine which of the subgropus of the symmetric group S(3) are normal. The condition is:

    for every g in G: g H g^-1 = H where H is a subgroup of G.

    I have determined all the subgroups of S(3) and I came up with 13.

    What I did after that is I considered 2 cases:

    1: g is elem. of H, then g^-1 is also in H and the condition is satisfied.
    2: g is not el. of H, but still el. of G, the same is true for g^-1.

    then I got stuck :(

    theoretically all the subgroups could be examined one by one, but it seems to me somehow too long and "not mathematical"

    Does anyone know a better way?

    thanks in advance!

    ° and another question: If H is a normal subgroup of G, what should one understand under "G modulo H"
  2. jcsd
  3. Nov 8, 2008 #2
    well, as you know:

    [tex] S_3=\{(1),(12),(13),(23),(123),(132)\}[/tex] Now since [tex]o(S_3)=6[/tex] it follows that its subgroups can be only of order 1, 2, 3 or 6. OF course the only subgroups of ord. 1 and 6 are the trivial one and S_3 itself respectively.

    So we are left with proper subgroups of ord 2 and 3. which are

    [tex] H_1=\{(1),(12)\}, H_2=\{(1),(13)\},H_3=\{(1),(23)\},H_4=\{(1),(123),(132)\}[/tex]

    So where did you get 13 subgroups?

    Then what i would do here, since the order of S_3 is not large, just look at the table for the operaton of S_3, and determine for which subgroups H_i, i=1,2,3,4=> gH=Hg. for every g in G.

    H_1, is not normal since, [tex] (123)^{-1}(12)(123)=(132)(12)(123)=(1)(23)=(23)[/tex] thus (23) is not in H_1.
  4. Nov 9, 2008 #3
    hmmm in my book the 6 elements are denoted:

    (123), (132), (213), (231), (312), (321) - how did you transform them in (1), (12), (13), (23), (123), (132)
  5. Nov 9, 2008 #4

    well what does (123) mean??

    It simply means the following, let f be the mapping then: f: 1-->2; 2-->3, 3-->1

    so following this logic, it yields that

    (123)=(231) =(312)and also (132)=(321)=(213) and these are actually only your even permutations in S_3, including (1) of course since (1)=(12)(21)

    Well, either you are misinterpreting what it is written in the book, or it is a typo, because the 6 elements of the symmetric group S_3 are the ones i listed.

    it is simply all the mappings that go from f:X_3--->X_3 where X_3={1,2,3} such that f is bijective.
  6. Nov 9, 2008 #5
    thanks, sutupidmath, I think I got it

    could you also help me eoth this: "If H is a normal subgroup of G, what should one understand under "G modulo H" ". I just cannot figure out what it means G modulo H in this context.
  7. Nov 9, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    If your textbook is defines normal subgroups it certainly should have a definition of G/H!

    A subgroup, H, of a group G is normal if and only if the right cosets, Hx, are the same as the left cosets, xH. That's important because it allows us to put an operation on the cosets: If xH and yH are two different left cosets of H, then "(xH)(yH)" is the left coset (xy)H. Of course, z is another member of xH, we can also represent that "zH". Would (zH)(yH)= (zy)H give the same coset? It can be shown that it will if and only if H is a normal subgroup of G- that the left and right cosets are the same.

    In that case, we define G/H to be the collection of all left cosets of H with that operation as group operation.
  8. Nov 9, 2008 #7
    but if G/H is a subgroup of G, for that xH=Hx for every x of [x], could we say then that G/H Abelian is?
  9. Nov 9, 2008 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    No, that does not follow.
  10. Nov 9, 2008 #9
    well I think I have to take a break now :)

    Thank you very much for the help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Normal groups
  1. Normal Groups (Replies: 15)