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Group theorem problems

  1. Jan 14, 2008 #1
    [SOLVED] group theorem problems

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

    1) suppose H is a finite subset of a group G. Prove that H is a subgroup of G if and only if xy belongs to H whenever x and y belong to H
    2) show that Q is not cyclic, (where Q is the group of rationals with addition.)

    3. The attempt at a solution

    1) if H is a subgroup, then if x,y are in H, so is by hypothesis xy, because H is a subgroup and therefore closed.

    Now conversely, if xy is an element of H when x,y are in H, then x*x is in H
    Let n be the number of elements in H, then x^n is in H
    Now because H is finite, there has to be an element such that x^n+1=x^m, with n+1>m. Otherwise there would be n+1 distinct elements in a set with n elements. I didn't get any further then this.

    I suppose if for elements of H a*b=a*c would imply that b=c, then x=x^n+1-m and so e:=x^n-m. Then I can find and inverse too without many problems, and because H is already know to have an associative operation, and H is closed by hypothesis, H is a subgroup. But is this true?

    2) Suppose there is an element x such that <x>=Q. Then x/2 is too in Q, and is not in <x>. But I don't find this a very convincing proof by contradiction.
     
    Last edited: Jan 14, 2008
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  3. Jan 14, 2008 #2

    CompuChip

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    1) Are you sure you aren't missing an inverse sign somewhere? As in: H is a subgroup of G iff for all x and y in H, [itex]x y^{-1} \in H[/itex] ? There was a case where you could do without the inverses (G abelian?), but in general you need it.

    2) Are Q the rationals, or the rationals - 0 or the quaternions or ...? A group consists of a set and an operation, what is the operation (addition, multiplication, ...).
     
  4. Jan 14, 2008 #3

    NateTG

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    In part 1, you can get a lot of mileage out of the fact that [itex]G[/itex] is a group, so every element of [itex]H[/itex] has some order in [itex]G[/itex].

    Compuchip: Since [itex]H[/itex] is finite, closure over multiplication is sufficient.
     
  5. Jan 14, 2008 #4
    compu: for 2) H is finite, which makes the difference. And Q are the rationals (with addition).

    nate, what exactly do you mean by 'all elements having some order in G', and what is it's use?
     
  6. Jan 14, 2008 #5

    CompuChip

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    You're right, I seem to be quite rusty in this subject.

    He means that for every element x in G, there is some number n such that x^n = e (not necessarily the same n for all x though).

    The only thing you know is that H is non-empty, so there must be at least one element, x. Now use what nate told you. You will be able to fulfill at least two of the group axioms for H in one blow.
     
  7. Jan 14, 2008 #6

    NateTG

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    Actually, [itex]G[/itex] can admit elements of infinite order since it is not necessarily finite. (Of course, those elements can't be part of [itex]H[/itex].)
     
  8. Jan 14, 2008 #7
    for 1) take any x in H, then x, x^2, x^3, ... are all elements of H, but H is finite, so at least two of these must be the same, say x^j = x^k for some j != k. You should be able to take it from here.
     
  9. Jan 14, 2008 #8
    ircdan, the problem with that argument is that you are not sure that xa=xb implies that a=b. But I think I got it now, thanks everyone!

    Anyone got any tips to show Q is not cyclic?
     
  10. Jan 14, 2008 #9
    It's almost ok! You need to rule out 0 at the beginning and this is obvious since <0> != Q. So suppose Q was cyclic, then there is x !=0 in Q s.t. <x> = Q. Then x/2 is in Q, but x/2 is not in <x>.
    This is enough, because <x> = {nx | n in Z} and x/2 is not of the form n*x for n in Z.

    Now if you are not convinced, suppose that x/2 was in <x>, then x/2 = nx for some n in Z.

    Then, x= (2n)x, so (2n-1)x = 0(ie, this means x + x + ... + x = 0 (2n-1) times), so x has finite order, but only 0 has finite order in Q, so x = 0, a contradiction.
     
    Last edited: Jan 14, 2008
  11. Jan 14, 2008 #10
    There is no problem there. If xa = xb in H, then this is an equation in G, so a = b. In our case everything is a power of x and x is in H, so everything is in H by hypothesis, so our inverse lies in H. It's a good trick to keep in mind!
     
  12. Jan 14, 2008 #11
    thanks, and solved!
     
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