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Proving stuff is a Subgroup

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Let a and b be integers

    (a) Prove that aZ + bZ is a subgroup of Z+
    (b) prove that a and b+7a generate aZ + bZ

    2. Relevant equations

    Z is the set of all integers


    3. The attempt at a solution

    (a)
    In order for something to be a subgroup it must satisfy the following 3 properties:

    (i)closure; that is that if aZ and bZ are in H (the subgroup of Z+) than aZ+bZ are in H.
    (ii) identity: 0 is in H
    (iii) inverses: if a(Z)+ b(Z) are in H, then so are -(a(Z) + b(Z)

    i really dont know how to prove any of these are true for this particular subset.

    I am also completley lost on part b.
     
    Last edited: Oct 12, 2008
  2. jcsd
  3. Oct 12, 2008 #2
    It's simpler than that: For H to be a subgroup of a group G, H must be nonempty and if a, b are elements of H, then ab-1 is an element of H.

    Start by proving that aZ + bZ is not empty. Then pick two arbitrary elements x and y in aZ + bZ and show that x - y is in aZ + bZ.

    Do you know what "a and b + 7a generate aZ + bZ" means?
     
  4. Oct 12, 2008 #3
    How do you show that aZ + bZ is non empty? it seems quite obvious to me but i dont know how to prove it.

    Is it something like:

    since a and b are integers, and when you mutiply integers with other integers, you get more integers.

    I think that a and b+7a generate aZ+bZ means something like a cyclic subgroup generated by the elements a and b+7a is aZ + bZ
     
  5. Oct 12, 2008 #4
    Give me a putative element of aZ + bZ and prove that it actually belongs to aZ + bZ.

    And how does that relate to this problem.

    Something like that. Can you provide more details.
     
  6. Oct 12, 2008 #5
    i know that the cyclic subgroup generated by a single element (lets say a) would be

    (... a^-2 , a ^-1 , I , a, a^2, a^3, ...)

    dunno how this works for 2 elements
     
  7. Oct 12, 2008 #6
    NM i figured this whole problem out

    Thanks.

    feel free to lock mods.
     
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