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Math Help - Groups and Subgroups

  1. #1
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    Groups and Subgroups

    Let Q be the set of rational numbers and H = { 5m/7^n | m, n are integers and n > o}.
    Determine whether H is subgroups of <Q, +>.
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  2. #2
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    Re: Groups and Subgroups

    Quote Originally Posted by mous99 View Post
    Let Q be the set of rational numbers and H = { 5m/7^n | m, n are integers and n > o}. Determine whether H is subgroups of <Q, +>.
    Verify (i) H\neq \emptyset. (ii) If m,n,m',n' integers with n>0,n'>0, then \dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q} with k,q integers and q>0.
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    Re: Groups and Subgroups

    Quote Originally Posted by FernandoRevilla View Post
    Verify (i) H\neq \emptyset. (ii) If m,n,m',n' integers with n>0,n'>0, then \dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q} with k,q integers and q>0.
    this is not enough. The last condition that remains is that every element in the subset has to have an additive inverse.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Groups and Subgroups

    Quote Originally Posted by rayman View Post
    this is not enough. The last condition that remains is that every element in the subset has to have an additive inverse.
    Those are necessary and sufficient conditions, according to a well known therorem (characterization of subroups).
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    Re: Groups and Subgroups

    Not according to my textbook. I am not sure what ''well known theorem'' you are referring to but according to subgroup criterion that is in my textbook (Durbin ''modern algebra'') in order to investigate whether a given subset is a subgroup we need to see it these criterions are satisfied
    1) set should not be empty
    2) it should be closed with respect to the operation
    3) every element of the set should have an inverse with respect to the operation
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    Re: Groups and Subgroups

    Rayman, it's commonly known as the one-step subgroup test. You'll notice that in Fernando's second statement, it has a minus, and not a plus.
    In any case, it's easy enough to see that the additive inverse will be from -m.
    i.e. \frac{5m}{7^n} + \frac{5(-m)}{7^n} = 0
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    Re: Groups and Subgroups

    and that completes the subgroup criterion
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Groups and Subgroups

    Quote Originally Posted by rayman View Post
    Not according to my textbook. I am not sure what ''well known theorem'' you are referring to but according to subgroup criterion that is in my textbook (Durbin ''modern algebra'') in order to investigate whether a given subset is a subgroup we need to see it these criterions are satisfied
    1) set should not be empty
    2) it should be closed with respect to the operation
    3) every element of the set should have an inverse with respect to the operation
    For if you are interested, here is proof that those three conditions are equivalent to only two: PlanetMath
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  9. #9
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    Re: Groups and Subgroups

    Quote Originally Posted by FernandoRevilla View Post
    Verify (i) H\neq \emptyset. (ii) If m,n,m',n' integers with n>0,n'>0, then \dfrac{5m}{7^n}-\dfrac{5m'}{7^{n'}}=\dfrac{5k}{7^q} with k,q integers and q>0.
    How to verify (i)?
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: Groups and Subgroups

    Quote Originally Posted by mous99 View Post
    How to verify (i)?
    Choose for example, m=0 and n=1. Then, \dfrac{5\cdot 0}{7^1}=0\in H.
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