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Finding a normal subgroup H of Zmn of order m

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a normal subgroup H of Zmn of order m where m and n are positive integers. Show that H is isomorphic to Zm.

    2. Relevant equations


    3. The attempt at a solution
    I am honestly not even sure where to start. My initial thoughts were if Zmn was isomorphic to Zm x Zn then I could find a subgroup H from that group. However, I discovered that Zmn is isomorphic to Zm x Zn but the converse is not true. Any help would be appreciated.

    Edit: If Zmn is cyclic has an element of order mn say x. Then nx has order m. Let H=⟨nx⟩.
    Now I just need to show that H is isomorphic to Zm, by constructing an isomorphism.
     
    Last edited: Mar 4, 2015
  2. jcsd
  3. Mar 4, 2015 #2

    Dick

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    You are just working with numbers mod mn here. Just put x=1. Think about the set of numbers {0,n,2n,3n,...,(m-1)n} like in your edit. Can't you see a correspondence with {0,1,2,3,...,(m-1)}?
     
    Last edited: Mar 4, 2015
  4. Mar 4, 2015 #3

    Stephen Tashi

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    What do you mean by that? Shouldn't you make an "If...then...." statement in order to speak of a converse?


    Instead of thinking about isomorphisms, you could think about a homomorphism and the "first isomorphism theorem" for groups. http://en.wikipedia.org/wiki/Isomorphism_theorem
     
  5. Mar 4, 2015 #4
    I do. However, I don't see how to construct a function based on this. I apologize for not being clear but the goal is to construct a bijective homorphic function. Then I can conclude that they are isomorphic.
     
  6. Mar 4, 2015 #5

    Dick

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    I would say try ##\phi(kn \pmod {nm})=k \pmod m## just from looking at it. Try to prove ##\phi## is an isomorphism.
     
    Last edited: Mar 4, 2015
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