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Question about Group theory

  1. Sep 30, 2008 #1
    If A and B are subgroups of G and AB=G does it follow that AB=BA? If yes, why?
  2. jcsd
  3. Sep 30, 2008 #2


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    I can't see why you can't have BA={1}.
  4. Sep 30, 2008 #3


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    tgt, can you give an example in which AB= G and BA= {1}?
  5. Sep 30, 2008 #4
    What does AB mean, i mean what operation are you performing between A and B here? is this supposed to be the same operation of the group G, or?
  6. Oct 1, 2008 #5


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    AB is the set of elements of the form ab (with a in A & b in B).
  7. Oct 1, 2008 #6


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    it doesn't mean I can see it.
  8. Oct 1, 2008 #7
    ok, so what about this.

    Like said AB is the set of all elements such a is in A and b is in B. Then from this follows that also [tex] a^{-1}b^{-1}\in AB, since, a^{-1}\in A, b^{-1}\in B[/tex] this comes from the fact that both A,B are subgroups , so they do have inverses.

    NOw since AB=G, it means also that the inverse of [tex] a^{-1}b^{-1}[/tex] is in AB. that is :

    [tex] (a^{-1}b^{-1})^{-1}=ba \in AB[/tex] So, from here we have:

    [tex] (a^{-1}b^{-1}) (a^{-1}b^{-1})^{-1}=e[/tex] or

    [tex](a^{-1}b^{-1}) ba=e[/tex] Now multiplying with a from the left side we get

    [tex]a(a^{-1}b^{-1}) ba=ae=>b^{-1}(ba)=a[/tex] now multiplying with b we get

    [tex] ba=ab[/tex] since ab, was chosen randomly from AB it means that AB=BA, doesn't it?

    I am not sure that this is absolutely right, someone else like halls or morphism, might comment on this!
  9. Oct 1, 2008 #8


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    Your proof should have stopped here - this is all you need to show. AB=BA is an equality of sets.

    As for the rest of your post:
    This step is wrong. Can you see why?
  10. Oct 1, 2008 #9
    Hell YES! This happens when you assume that the same thing is gonna happen in next step as well. We cannot do this, because we dont know whether A, B are commutative. so by doing what i did in the last step, i should have multiplied by b from the left side, which would get us nowhere, since we would end up with ba=ba... and this is not what we wanted.

    Well, i first also thought that i should stop here
    [tex] (a^{-1}b^{-1})^{-1}=ba \in AB[/tex]

    but, like i said, wasn't sure enought, so i kept goin, but if i would have noticed earlier that my last step was nothing productive, then i probbably would have realized that this
    [tex] (a^{-1}b^{-1})^{-1}=ba \in AB[/tex] is all we need.. which also makes sens, sice this shows that for every element ab in AB, where a is in a,and b is in be, also ba is in AB, which means that b is in A, and b is in B. now sice, like said, ab were randomly chosen, that is for any ab, from AB it means that AB=BA.
    Last edited: Oct 1, 2008
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