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Proof about a group

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Let G be an ableian group of order mn, where m and n are relativiely prime. If G has
    has an element of order m and an element of order n, G is cyclic.
    3. The attempt at a solution
    ok so we know there will be some element a that is in G such that
    [itex] a^m=e [/itex] where e is the identity element. It seems that this would be enough to prove that their is a sub group generated by a. and this sub group is cyclic. if I start with the element a
    all powers of a would need to be in their so it would be closed under the operation.
    I guess we know its a group already. Lets say we have some power of a like x where
    0<x<m we want to know if this has an inverse that is a power of a.
    we know [itex] a^m=e [/itex] so if we have some arbitrary power of a [itex] a^x [/itex]
    we want its inverse [itex] a^xa^p=e=a^{x+p}=a^m [/itex] so x+p=m so their is a cyclic subgroup
    generated by a, Now we know that if we have a cyclic group all of its subgroups are cyclic.
    I am slightly worried about the converse, is it always true if I have cyclic subgroup that the group is cyclic? But I guess i could just do the same argument with some element of the form
    [itex] b^n=e [/itex] and then look at all the possible group operations. I guess I could try to find the generator for G.
     
  2. jcsd
  3. Apr 8, 2013 #2

    jambaugh

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    What is the order of ab?
     
  4. Apr 8, 2013 #3
    mn, so I guess ab would be the generator of the group.
     
    Last edited: Apr 9, 2013
  5. Apr 9, 2013 #4

    jambaugh

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    Well, you must argue that. Given its order and the size of the group, what's left?
     
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