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Abstract Algebra: Proving whether H is a subgroup.

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Let R = {all real numbers}. Then <R,+> is a group. (+ is regular addition)
    Let H = {a|a [tex]\epsilon[/tex] R and a2 is rational}.

    Is H closed with respect to the operation?
    Is H closed with respect to the inverse?
    Is H a subgroup of G?

    2. Relevant equations
    N/A

    3. The attempt at a solution

    For the inverse:

    From my notes I know that If it is closed in respect to the inverse that I'll have to prove:
    If x[tex]\epsilon[/tex] H, then x inverse [tex]\epsilon[/tex] H. If it's not I'll have to show an element b [tex]\epsilon[/tex] H, but b-1 is not an element of H.
    I know for the inverse that a.1 = -a for addition. It's looking like it's yes since a and -a is and element of R but I have to prove that somehow and I'm not sure how.

    For the operation:

    My notes show for operation that H is closed in respect to the operation if a * b [tex]\epsilon[/tex] H for any a,b [tex]\epsilon[/tex] H.

    Knowing that I wrote a+b [tex]\epsilon[/tex] H for any a,b [tex]\epsilon[/tex] H. Now I'm pretty sure that this is also yes but I'm going to have to prove if x,y [tex]\epsilon[/tex] H, then x +y [tex]\epsilon[/tex] H and I'm not sure how to actually prove it.

    I know that if both of these is 'yes' then it is indeed a subgroup.

    Help is greatly appreciated.
     
    Last edited: Feb 17, 2010
  2. jcsd
  3. Feb 17, 2010 #2

    CompuChip

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    First a general remark: instead of showing that a non-empty subset H of a group G satisfies all the axioms, it is sufficient to show that [itex]x y^{-1} \in H[/itex] for all x and y in H.

    For the first question, how about [itex]a = \sqrt{2}, b = \sqrt{3}[/itex] ?

    For the second one, just use the definition. Suppose a is in H, then a2 is rational. Is (-a) in H then, i.e. is (-a)2 rational?
     
  4. Feb 17, 2010 #3
    Ok good. First to clarify notation, is it the inverse of x time y or is it x times the inverse of y?
    How would I go about showing that?


    So this is a counterexample that will prove it isn't an operation?

    Oh I understand. How would you actually recommend proving that though?
     
  5. Feb 17, 2010 #4

    CompuChip

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    Sorry, it means x times the inverse of y.
    It's a good exercise in group theory to prove that statement (or look up the proof and follow it closely).


    About your last question... take another look. We know that a2 is rational. How about (-a)2 ?
     
  6. Feb 18, 2010 #5
    If I did this correctly I was wrong for both of them. :rofl:

    For the operation I did:

    [tex]\sqrt{2}[/tex], [tex]\sqrt{3}[/tex] [tex]\in[/tex] H but [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] is not in H.

    Reason: [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] is a real number but , [tex]\left(\sqrt{2}+\sqrt{3}\right)^{2}[/tex] is not rational.
    Therefore H is not closed with respect to the operation.


    For the inverse I did:

    H is closed with respect to the inverse:

    Suppose a is in H then a[tex]^{2}[/tex] is rational. If -a is in H, then (-a)[tex]^{2}[/tex] is rational.

    Since 0 is the identity, a + (-a) = 0
    So if 'a' is [tex]\in[/tex] H, a[tex]^{-1}[/tex] [tex]\in[/tex] H

    and since H isn't closed with respect to both the inverse and the operation it is not a subgroup of G.

    How's that?
     
  7. Feb 18, 2010 #6

    CompuChip

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    Agreed, although it would be nice if you showed this explicitly. For example, if you'd take [itex]a = b = \sqrt{2}[/itex] then (a + b)2 would be rational...

    This is still a bit ... fuzzy. All your statements are true, but where do you prove that (-a) is in H? I.e. where can I read that (-a)2 is rational?
     
  8. Feb 18, 2010 #7
    Oh. ok.

    Oh that's right. :rofl:

    I should have said that:
    For <R,+>: a+0=a thus making 0 the identity.
    then a+x=0 for determining the inverse.
    thus -a would be the the inverse for the operation.

    I'm not sure where to go from there.
     
  9. Feb 18, 2010 #8

    vela

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    You're getting sidetracked proving what you already know. You're given that (R,+) is a group. That means you already know that -a is the additive inverse and 0 is the additive identity. The only question you need to answer is: if a is in H, is -a also in H?
     
  10. Feb 18, 2010 #9
  11. Feb 23, 2010 #10

    Yes. That's actually where I'm stuck right now. I need to show why -a is in H.

    I have that

    Since for <R,+> e = 0, then a +a-1 = 0 and therefore a-1= -a

    What I need to show now is that -a is in H and I don't know how to begin doing that.

    For the second part:

    For the operation, I need to show a counter example to show that it is not closed in respect to the operation.

    I have to show a counter example that shows that it is not rational and therefore not in H. I had [tex]\sqrt{2}[/tex] and [tex]\sqrt{3}[/tex] being added together and then squared to show it but it doesn't show why. Any ideas on what I can do?

    Both of these concerns are open to anyone who can help.
     
    Last edited: Feb 23, 2010
  12. Feb 23, 2010 #11

    vela

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    What's the criterion for whether any element is in H? Does -a satisfy that criterion?

    What do you mean it doesn't show why?
     
  13. Feb 24, 2010 #12

    CompuChip

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    Hey The_Iceflash,

    When I read your replies, I begin to suspect that the problem lies in your understanding of the definition. Can you tell me whether the following numbers are or are not elements of H?

    [tex]2, \sqrt{2}, \sqrt{\sqrt{2}}, \pi^3, \frac{4}{3}, \frac{\sqrt{3}}{2}, \left( \frac{1}{7} \right)^{1/3}[/tex]
     
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