Show that nZ intersection mZ= lZ

  1. hello

    i have two questions and i need answers for them

    first one:

    in the additive group (Z,+)
    show that nZ intersection mZ= lZ

    , where l is the least common multiple of m and n.

    The second question is :

    Given H and K two subgroups of a group G , show the following:

    (H union K) subgroup of G if and only if H subset of K or K subset of H

    :confused: :smile:
  2. jcsd
  3. matt grime

    matt grime 9,395
    Science Advisor
    Homework Helper

    What have you done and what do you think you need to do? Remember, the first thing to do (if you can't do the question straight away) is to write down the definitions of everything so you know what you have to do.
    Last edited: Apr 21, 2006
  4. HallsofIvy

    HallsofIvy 41,269
    Staff Emeritus
    Science Advisor

    You're not going to get answers! Show us what you have done and, as precisely as you can, where you have difficulty and you will get help
  5. [​IMG]

    Is this solution true for question two ??
  6. matt grime

    matt grime 9,395
    Science Advisor
    Homework Helper

    You need to learn to communicate by posting the maths here not posting a link to a zipped file with two jpgs in it.

    It has the seed of a correct solutin, but no, I think it is not correct. You say:

    let x be in H and Y be in K and then show if xy is in H y is in H and then conclude that K C H, and dually that if xy is in K then x isin K and H C K. But that doesn't follow. If for ALL x and y in H and K respectively then either xy is always in H or xy is always in K and it would follow, but there is no reason to suppose this at all.

    Suppose that neither H C K nor K C H, then there are elements x in H but not in K and y in K but not in H. Where is xy? (I.E. apply your argument now to get a contradiction.)

    Perhaps better is to do: if H C K we're done, if not then (and apply your argument to well chosen x in H).

    You also should take care between using < and C, subgroup and subset.
    Last edited: Apr 23, 2006
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