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Homework Help: Subgroup math properties help

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Let G be a group and let [tex]A \leq G[/tex] be a subgroup. If [tex]g \in G[/tex], then [tex]A^g \subseteq G[/tex] is defined as

    [tex]A^g = \{ a^g | a \in A \}[/tex] where [tex]a^g = g^{-1}ag \in G[/tex]

    Show that Ag is a subgroup of G.

    3. The attempt at a solution

    I will use the one step subgroup test. First I have to identify the property that distinguishes the elements of Ag (a defining condition). I don't see what's the binary operation so I can't tell what this property is... :confused:

    If I knew this property, I would prove that the identity has this property, so that Ag is nonempty. Then I'd use the assumption that a and b have that property to show that ab-1 has this property. Could anyone help me out to see what the property of this group is?

    P.S. I think the identity for this group is "1", right?
  2. jcsd
  3. Aug 1, 2010 #2


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    Re: Subgroups

    isn't it given by: [itex]a^g = g^{-1}ag \in G[/itex]?

    then from memory you need to show:
    - the identity is part of the subgroup
    - its closed under multiplication, ie if u & v are both in, then so is uv
    - the inverse of any element is also part of subgroup

    these whould all follow pretty straight forwarly from the group definition, remembering that A is a subgroup itself
  4. Aug 2, 2010 #3
    Re: Subgroups

    I want to use the "one step subgroup test". For this test I don't have to check closure. I only need to check that ab-1 is in the group whenever a and b are in the group.

    To show that identity is in the group is it sufficient to show that g-11g=1? This verifies that Ag in nonempty.

    Suppose a=g-1ag and b=g-1bg are elements in Ag who satisfy the property.

    b-1 = gbg-1

    ab-1 = g-1ag . gbg-1

    What can I do next? :rolleyes:
  5. Aug 2, 2010 #4


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    Re: Subgroups

    No, this is wrong. You can show that [itex](gbg^{-1})^{-1}= gb^{-1}g^{-1}[/itex] directly from the definition of "inverse".

  6. Aug 3, 2010 #5
    Re: Subgroups

    I made some mistakes, so I will start over.

    We write two elements of Ag in the correct form (correct form here being [tex]a^g = g^{-1}ag[/tex]):

    ag and bg

    Then I must show that [tex]a^g(b^g)^{-1}[/tex] also has the correct form; that is it can be written as [tex]g^{-1}(ab^{-1})g[/tex]. I start here:

    [tex]a^{g}(b^{g})^{-1} = (g^{-1}ag)(gb^{-1}g^{-1})[/tex]

    But how does one simplify [tex](g^{-1}ag)(gb^{-1}g^{-1})[/tex] to get [tex]g^{-1}(ab^{-1})g[/tex]? This is where I'm stuck...
  7. Aug 3, 2010 #6


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    Re: Subgroups

    why do you assume [itex](b^{g})^{-1} = (gb^{-1}g^{-1})[/itex] ?
  8. Aug 3, 2010 #7


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    Re: Subgroups

    cos its not... try multiplying with [itex]b^{g}[/itex]
  9. Aug 3, 2010 #8
    Re: Subgroups

    Because [tex]b^g = g^{-1} b g[/tex]

    therefore [tex](b^g)^{-1}= (g^{-1} b g)^{-1} = g b^{-1} g^{-1}[/tex]

    could you show me? because I don't understand what you mean here...
  10. Aug 3, 2010 #9
    Re: Subgroups

    just remember you need to switch orders when you invert (ab)-1 = b-1 a-1.
  11. Aug 3, 2010 #10


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    Re: Subgroups

    or just start with b^g and just see what you would have to multiply it by to get the identity
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