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Question about Cyclic groups

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data
    My online notes stated that it |g| = |G| where g is an element of G then |G| is cyclic.

    Can somebody help me understand why this is true?
     
  2. jcsd
  3. Aug 5, 2014 #2

    jbunniii

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    I assume ##|g|## means the order of ##\langle g \rangle##, the subgroup generated by ##g##.

    Note that ##\langle g \rangle \subseteq G##.

    If ##|G|## is finite and ##|g| = |G|##, then ##\langle g \rangle \subseteq G## implies ##\langle g \rangle = G##. (Do you see why?) What can you conclude?

    By the way, the result need not be true if ##|G|## is infinite. For a counterexample, let ##G = \mathbb{Q}##, the additive group of rational numbers, and let ##g = 1##. Then ##\langle g \rangle = \mathbb{Z}##, the additive subgroup of integers. Then ##G## and ##\langle g \rangle## have the same cardinality (countably infinite) but ##G## is not cyclic.
     
  4. Aug 5, 2014 #3

    pasmith

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    If [itex]|g| = |G|[/itex] is finite, can there exist any elements of [itex]G[/itex] which are not powers of [itex]g[/itex]?
     
  5. Aug 5, 2014 #4

    HallsofIvy

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    No. If |g|= |G| then, by definition of "| |", g and G contain the same number of terms. Since <g> is always a subgroup of G, it follows that g is exactly the same as G. Since every member of <g> is a power of g, every member of G is a power of g.
     
  6. Aug 5, 2014 #5
    thanks yall :D
     
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