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I Nature of cyclic groups

  1. Feb 21, 2017 #1
    We define a cyclic group to be one all of whose elements can be written as "powers" of a single element, so G is cyclic if ##G= \{a^n ~|~ n \in \mathbb{Z} \}## for some ##a \in G##. Is it true that in this case, ##G = \{ a^0, a^1, a^2, ... , a^{n-1} \}##? If so, why? And why do we write a cyclic group as ##\{a^n ~|~ n \in \mathbb{Z} \}##, where n is allowed to be any integer, except just those 0 through n - 1?
     
  2. jcsd
  3. Feb 21, 2017 #2

    fresh_42

    Staff: Mentor

    Yes. And until now, nothing is said about the order of ##a##. It can be of finite order or of infinite order.
    In the first case, say ##a^m=1## we get ##(G,\cdot) \cong (\mathbb{Z}_m,+)## and in the second case ##(G,\cdot) \cong (\mathbb{Z},+)##
    Where's the rest of the sentence? Is it true that in case ##G \cong \mathbb{Z}_n## ... what?
    This doesn't make sense. There is no restriction on ##n## as it runs through all integers. The only question is, whether ##a^m=1## for some integer ##m## or not. If ##a^m=1## then ##m=0## will get us ##G=\{1\}~,## if ##m<0## then we can exchange ##a## by ##a^{-1}## (which is in the group and also a generator) and get an ##m > 0##, so that we may always assume ##m \in \mathbb{N}##. If there is no such number ##m##, then ##G## is a free Abelian group of rank ##1## which is isomorphic to ##\mathbb{Z}~##.
     
  4. Feb 21, 2017 #3

    Math_QED

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    We can only write it that way when G is finite.
     
  5. Feb 22, 2017 #4
    Assume that G is a finite cyclic group. What I'm asking is whether writing ##G = \{a^n ~ | ~ n \in \mathbb{Z} \}## is the same as writing ##\{a^0, a^1, ..., a^{n-1} \}##, and if so, why?
     
  6. Feb 22, 2017 #5

    Math_QED

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    Yes: (I used o(g) to denote the order of the element g)

    Proof:

    If G is cyclic, we know there is an element ##g \in G## such that ##G = \{g^k|k \in \mathbb{Z}\}##. We also know that ##g^i = g^j \iff i \equiv j (mod \quad o(g))##.
    Assume ##o(g) = m##. As ##G## is finite, ##m## is finite as well ##(m \in \mathbb{N}_0)##. Then, we see that ##G = \{e,g,g^2,g^3, ..., g^{m-1}\}##, because the other elements were duplicates of those elements (as we can reduce the exponents modulo m; for example: ##g^m = g^{2m}=g^{-m} = e, g^2 = g^{2-m} = g^{m+2}##). Every two elements that are still left are different (as they cannot be reduced mod m and give another element in the list).

    I put the key part where you need that ##G## is finite in italics.

    EDIT: Interesting to note:

    Consider this theorem:

    Let ##G## be finite of order ##n##. Then:

    ##G## cyclic ##\iff \exists g \in G: o(g) = n##
     
    Last edited: Feb 22, 2017
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