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Operations on a group

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    An operation of a group G on a set S is a function G X S [tex]\rightarrow[/tex] S satisfying:

    1. es = s [tex]\forall[/tex]s [tex]\epsilon[/tex] S
    2. g(hs) = (gh)s [tex]\forall[/tex]g,h [tex]\epsilon[/tex] G, s[tex]\epsilon[/tex] S

    If s [tex]\epsilon[/tex] S, show that the stabilizer of s, defined as the set:
    {g [tex]\epsilon[/tex] G | gs = s}
    is a subgroup of G

    2. Relevant equations



    3. The attempt at a solution

    Well from that definition it seems that g must be the identity element of G. Is a set consisting of just the identity not just a group? And since G is a group it includes the identity, thus the set g: {e} is a subgroup of G?
     
  2. jcsd
  3. Nov 6, 2009 #2

    lanedance

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    i'm not sure about that...

    my group theory isn't the best, but... i think the identity will surely be a part (from the defintion), but also potentially other elements as well

    for example consider the group of all 2D rotations, i'm not 100% sure how to put it together, but imagine the set as a square, defined only by the location of its 4 vertices, only rotations of n*(pi/2) take the square into itself... i think the the stabiliser group in this case, then represents the rotational symmetry group of the square...

    so i think the key is to show something along the lines of:
    - the identity must be in the stabiliser group, and if f,h are in the stabiliser group, so is fh & h^{-1}
     
    Last edited: Nov 6, 2009
  4. Nov 6, 2009 #3

    lanedance

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    once again expanded above
     
  5. Nov 6, 2009 #4

    HallsofIvy

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    No, the set of all g obviously contains e but there may be other members of G that "fix" s. Remember that s is NOT itself a member of G. You can only say that es= s because of (1) in your definition of the action of G on S.

    (1) tells you that e is in this set and (2) tells you that the operation is associative. Now you need to prove that this set is closed under the operation: if gs= s and hs= s, then (gh)s= s. You also need to prove that if gs= s, then g-1s= s.
     
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