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When are groups finite?

  1. Jun 27, 2010 #1
    Is it true that cyclic groups with [tex]x^n = 1[/tex] the only finite groups (with order n)?

    I've been experimenting with a few groups and I think this is true but I'm not sure.


    thanks
     
  2. jcsd
  3. Jun 27, 2010 #2

    CompuChip

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    Have you looked at the permutation groups Sn, dihedral groups Dn (symmetries of regular polygons), etc.?

    In fact, almost all finite groups are non-abelian, so the cyclic groups are just very few in number.
     
  4. Jun 27, 2010 #3

    phyzguy

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    The finite groups are completely categorized. See for example:

    http://en.wikipedia.org/wiki/List_of_finite_simple_groups

    Or, I strongly recommend Mark Ronan's book, "Symmetry and the Monster". It is a good read of this fascinating subject.
     
  5. Jun 28, 2010 #4
    There are as many cyclic groups ([itex]\aleph_0[/itex]) as finite groups. By, "almost all finite groups are non-abelian", do you mean [itex]\lim_{n\rightarrow \infty}a_n/g_n=0[/itex], where [itex]a_n[/itex] is the number of abelian groups of order [itex]\leq n[/itex] and [itex]g_n[/itex] is the number of groups of order [itex]\leq n[/itex]? (This doesn't seem likely.)
     
  6. Jun 28, 2010 #5

    HallsofIvy

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    Given any n, there certainly exists a cylic group of order n.

    It is also true that if p is a prime number then then the only group of order p is the cylic group.

    But if n is NOT prime, then there exist other, non-cyclic, groups of order n.

    The simplest example is for n= 4. The "Klein 4-group" is not cyclic.

    The Klein 4-group has 4 members, e, a, b, c satifying
    ee= e, ea= a, eb= b, ec= c (e is the identity)
    ae= a, aa= e, ab= c, ac= b
    be= b, ba= c, bb= e, bc= a
    ce= c, ca= b, cb= a, cc= e

    The fact that every element is its own inverse proves this group is not cyclic.

    Any group of order 4 is isomorphic either to the cyclic group of order 4 or to the Klein four-group.
     
  7. Jun 28, 2010 #6
    Then again it doesn't seem unlikely. Has it been proved?

    The non-abelian groups of orders 2n will probably arrange it by themselves (e.g. there are over ten times as many non-abelian groups of order 1024 as there are other groups up to and including 1024) so this may not be too difficult to prove.
     
    Last edited: Jun 28, 2010
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