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Prove product of infinite cyclic groups not an infinite cyclic group

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the product of two infinite cyclic groups is not an infinite cyclic?

    2. Relevant equations

    Prop 2.11.4: Let H and K be subgroups of a group G, and let f:HXK→G be the multiplication map, defined by f(h,k)=hk.

    then f is an isomorphism iff H intersect K is {1}, HK=G, and also H and K are normal subgroups of G.

    3. The attempt at a solution

    Here is the outline of my proof. It didn't match a lot of things I saw online so I figured I would ask if my logic was ok.

    Proof by contradiction

    1.)Let Cm and Cn be an infinite cyclic groups.
    2.) Assume CmXCn is isomorphic to Cm (or Cn or anything other infinite cyclic group I think.)

    I felt least comfortable with this step. My reasoning for this step though is that all infinit cyclic groups are isomorphic to to the the integers under addition. This if cm and cn are isomorphic to the same thing then they must be isomorphic to each other.

    3.)Since CmXCn is an isomorphism to Cm then CmCn=Cm. This would imply that Cn={1} but this would be a contradiction since Cn is supposed to be infinite.

    Thank you for your time.
  2. jcsd
  3. Nov 6, 2011 #2


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    You are off on the wrong track with the theorem and your 'proof' doesn't have much substance. Look, if H and K are your infinite cyclic groups, then the operation on HxK is (h1,k1)*(h2,k2)=(h1*h2,k1*k2). If HxK were cyclic it would have to have a single generator. Think about it a bit more.
  4. Nov 6, 2011 #3


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    it is sufficient, i believe, to show ZxZ is not cyclic.

    what are the only candidates for a generator?
  5. Nov 7, 2011 #4


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    Sure it is. That's what I meant by 'think about it some more'. It's not that hard a problem.
    Last edited: Nov 7, 2011
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