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Abstract Algebra Question

  1. Feb 1, 2012 #1
    Show that [itex]\langle[/itex] a,b [itex]\rangle[/itex] = [itex]\langle[/itex] a,ab [itex]\rangle[/itex] = [itex]\langle[/itex] a^-1,b^-1 [itex]\rangle[/itex] for all a and b in a group G


    I am not sure what this question is asking. Does this notation mean that a the cyclic group is generated by a,b and any combination of the two?
     
    Last edited: Feb 1, 2012
  2. jcsd
  3. Feb 1, 2012 #2

    Dick

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    I don't think they said the group was cyclic and I don't think they said that any of those pairs generate the whole group. I think you are supposed to show they generate the same subgroup of G.
     
  4. Feb 1, 2012 #3

    Deveno

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    one way to show that <S> = <T> (where S and T are subsets of G), is to show every element of T is in <S>, and vice-versa.
     
  5. Feb 1, 2012 #4
    ok but what does[itex]\langle[/itex] a,b [itex]\rangle[/itex] actually mean. I really dont understand what I am suppose to do.
     
  6. Feb 1, 2012 #5

    Deveno

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    [tex]\langle a,b \rangle[/tex]

    means the subgroup of G generated by a and b. this is, by definition, the smallest subgroup of G containing the set {a,b}, and will contain as subgroups the cyclic group generated by a, and the cyclic group generated by b.

    you will have to use the fact that if a is in a (sub)group (as a generator, or product, or whatever), then so is a-1.

    for example, it's pretty clear that ab is in the subgroup generated by a and b (ab is in <a,b> = <S>, where S = {a,b}), because ab is in this subgroup (by closure). the trick is, can you show that b is in the subgroup generated by a, and ab?
     
  7. Feb 1, 2012 #6
    Would I say something like

    a*[itex]\langle[/itex] a,b [itex]\rangle[/itex]=[itex]\langle[/itex] a^2,ab [itex]\rangle[/itex]

    but since a generates a^2 we can leave it as a. Or would this be wrong?
     
  8. Feb 1, 2012 #7

    Dick

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    Wrong. Another way to look at <a,b> is that it is the set of all finite products of powers of a and b. Like a^3*b^(-1)*a^6*b^(-3). Etc etc etc. Change the powers and number of a's and b's anyway you like. Now pay more attention to what Deveno said in post 3.
     
    Last edited: Feb 1, 2012
  9. Feb 1, 2012 #8
    So since a and b generate some subgroup I can just say that a, ab must generate the group because any combination of the two elements will generate the subgroup
     
  10. Feb 1, 2012 #9

    Dick

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    That is so vague as to be almost meaningless. Read Deveno's post again. Deveno said something pretty specific that you seem to be missing.
     
  11. Feb 1, 2012 #10
    he said there are two subgroups that are equal so they are the same subgroup...
     
  12. Feb 1, 2012 #11

    Dick

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    This one. E.g. show every element of {a,ab} is in <a,b> and show every element of {a,b} is in <a,ab>.
     
  13. Feb 1, 2012 #12
    ok, how about this
    Consider the subgroup generated by <a,b>. Clearly a is in <a,b> and since a sub group is closed we know that ab is in <a,b>.

    On the other hand consider the subgroup generated by <a,ab> since <a,ab> is a sub group its clear a is in <a,ab>. Also there exist and a^-1 so a^-1(ab) is in <a,ab> this implies that b is in<a,ab>.


    Is that right?
     
  14. Feb 2, 2012 #13

    Dick

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    Yes, I think you are getting the idea.
     
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