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For every positive integer n there is a unique cyclic group of order n

  1. May 3, 2014 #1
    Hi,

    I can't understand why the statement in the title is true. This is what I know so far that is relevant:

    - A subgroup of a cyclic group G = <g> is cyclic and is <g^k> for some nonnegative integer k. If G is finite (say |G|=n) then k can be chosen so that k divides n, and so order of g^k is n/k.

    - Let G = <g> be a cyclic group. If g has infinite order, then G is isomorphic to the integers with addition. If o(g) = n, then G is isomorphic to Z/nZ with addition.

    And then apparently the last point implies the statement in my title and I have no idea how!

    Thanks,

    Matt
     
  2. jcsd
  3. May 3, 2014 #2

    Fredrik

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    I'm not entirely familiar with the standard terminology (or this theorem), so you will have to explain a few things. The order of a group is the cardinality of its underlying set, right? But what do you mean by order of g? Isn't g an element of the group? Was that supposed to be "order of G"? What do you mean by <g>? I would guess that it's the smallest subgroup that has {g} as a subset. What does o(g) mean? Is it "order of g", where "g" actually means G? In that case, I'd say that the "last point" (the sentence that contains o(g)) is just a slightly stronger statement than the one in the thread title, and there's nothing to prove.

    I assume that "unique" means "unique up to isomorphism".
     
  4. May 3, 2014 #3

    jbunniii

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    I agree with Fredrik that there seems to be nothing to prove, if you already have the quoted fact:
    In other words, up to isomorphism, there is exactly one cyclic group of order ##n##, namely ##\mathbb{Z}/n\mathbb{Z}##.

    @Fredrik - you are correct about the meanings of all of the terms, which are all standard usage in group theory:

    Order of ##G## = cardinality of the underlying set

    ##\langle g \rangle## = subgroup generated by ##g## = smallest subgroup of ##G## which contains ##g##. This is always a cyclic group consisting of the elements ##g^n## for all integers ##n##. If the powers are all distinct, then we have an infinite cyclic group, otherwise there is some smallest positive ##n## such that ##g^n## is the identity. In this case, there are ##n## distinct elements and the group is isomorphic to ##\mathbb{Z}/n\mathbb{Z}##.

    ##o(g)## = order of ##g## = order of ##\langle g \rangle## ("order" is overloaded here, but the meaning is always clear from the context)
     
    Last edited: May 3, 2014
  5. May 3, 2014 #4
    What exactly does 'up to isomorhism' mean, and why does this mean there isn't more than one?
     
  6. May 3, 2014 #5

    Fredrik

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    It means that there may be infinitely many, but they're all isomorphic to each other.
     
  7. May 3, 2014 #6

    Erland

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    It means that all cyclic groups of order n are isomorpic to each other.
     
  8. May 3, 2014 #7
    Ah! That makes sense now. Thanks very much!
     
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