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Cyclic groups generator problem!

  1. Jan 30, 2012 #1
    Regarding finite cyclic groups, if a group G, has generator g, then every element [itex] h \in G [/itex] can be written as [itex] h = g^k [/itex] for some k.

    But surely every element in G is a generator as for any [itex] k [/itex], [itex] (g^k)^n [/itex] eventually equals all the elements of G as [itex] n [/itex] in takes each integer in turn.

    Thanks for any replies!
  2. jcsd
  3. Jan 30, 2012 #2


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    This is not true. What if you choose a k such that g^k=1 (e.g. k=|G|)?

    The problem with your reasoning is that (g^k)^n=g^(kn) won't hit every element of G as n runs, because you need all elements of the form g^m for any m. If you're only taking powers of the form kn then you won't get every integral power unless you choose a good k. It's pretty easy to determine what makes k good. You simply need to recall two facts:
    (1) ord(g^k)=ord(g)/gcd(ord(g),k)=|G|/gcd(|G|,k), and
    (2) h generates G iff ord(h)=|G|.
    Combining these, we conclude that g^k will generate G=<g> iff gcd(|G|,k)=1, i.e., iff k and |G| are coprime.

    So my example above with k=|G| (or more generally k a multiple of |G|) is an example of the worst possible k to choose.
  4. Jan 30, 2012 #3
    Ah I see it now, thank you!
  5. Jan 31, 2012 #4


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    just try adding 2 to itself repeatedly in Z/6Z. note that you only get different results for n= 1,2,3.
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