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Question on cyclic groups

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]G[/itex] be a cyclic group of order [itex]n[/itex] and let [itex]k[/itex] be an integer relatively prime to [itex]n[/itex]. Prove that the map [itex]x\mapsto x^k[/itex] is sujective.


    2. Relevant equations



    3. The attempt at a solution
    I am trying to prove the contrapositon but I am not sure about one thing: If the map is not surjective, does it necessarily mean that there exists distinct [itex]i,j \in \{1,...n\}[/itex] such that [itex](x^i)^k=(x^j)^k[/itex]? If so how would you prove it?

    Anyway here is my proof:

    Suppose that the map is not surjective. Then there exists distinct [itex]i,j \in \{1,...n\}[/itex] such that [itex](x^i)^k=(x^j)^k[/itex]. Without loss of generality suppose [itex]i>j [/itex]. Using the cancellation laws we get [itex]x^{(i-j)k}=1[/itex]. Since [itex]|x|=n[/itex], it follows that [itex]n|(i-j)k[/itex] (By another proposition). If [itex]gcd(n,k)=1[/itex] then [itex]n|(i-j)[/itex], a contradiction since [itex](i-j) < n[/itex]. Hence we must have [itex]gcd(n,k) \not= 1[/itex] and so [itex]k[/itex] is not relatively prime to [itex]n[/itex]. Therefore by contraposition, if [itex]k[/itex] is relatively prime to [itex]n[/itex] then [itex]x\mapsto x^k[/itex] is surjective.

    Quite often I find it hard to check whether a proof has flaws in it. How can I improve on checking for flaws in a proof?

    Any help would be appreciated.
     
  2. jcsd
  3. Jan 22, 2012 #2

    Dick

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    Your proof looks fine to me. If your group is cyclic and x is a generator then your group is G={x^0,x^1,...x^(n-1)}. So, yes, if x->x^k is not surjective then two of those elements must map to the same thing. Hence (x^i)^k=(x^j)^k for some i and j less than n. Sometimes putting real numbers in for n and k helps to check, e.g. put n=6. Show the map is surjective if k=5 and not surjective if k=4 by writing all of the elements out. It should give you a feeling for what's going on if the proof itself isn't giving you that.
     
  4. Jan 23, 2012 #3
    Thanks for the reply Dick.

    So the image of the map is a subset of G, that is why if the map is not surjective then two elements must map to the same thing. Is that correct?
     
  5. Jan 23, 2012 #4

    Dick

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    Sure. If x is in G, then x^k is in G. Groups are closed under the operation.
     
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