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Fundamental Theorem of Abelian Groups

  1. Sep 28, 2011 #1
    Show that there are two abelian groups of order 108 that have exactly one subgroup of order 3.

    108 = 2^ 2 X 3 ^ 3

    Using the fundamental theorem of finite abelian groups, we have

    Possible abelian groups of order 108 can be : Z108, Z4 + Z27, Z2+Z2+Z27, Z4+Z9+Z3, Z2+Z2+Z9+Z3, Z4+Z3+Z3+Z3, Z2+Z2+Z3+Z3+Z3

    It seems to me that all three Z108, Z4+Z27, Z2+Z2+Z27 have exactly one subgroup of order 3. Please suggest where am I going wrong ?
  2. jcsd
  3. Sep 28, 2011 #2
    The groups [itex]\mathbb{Z}_{108}[/itex] and [itex]\mathbb{Z}_4\times \mathbb{Z}_{27}[/itex] are isomorphic.
  4. Sep 28, 2011 #3
    Thanks, Micromass.
  5. Sep 28, 2011 #4
    Sorry to nitpick, mehtamonica: You may be using shorthand, but I think that should be finitely-generated Abelian groups, not finite Abelian groups.
  6. Sep 28, 2011 #5


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  7. Sep 29, 2011 #6
    Well, in the OP, in line 3, Mehtamonica referred to the' fundamental theorem of

    finite Abelian groups' ; Z -integers is clearly not finite; so it is the FT of fin.gen.

    Abelian groups.
  8. Sep 29, 2011 #7
    Yes, it is usually phrased as finitely generated, although of course it will imply as a corollary that all finite ones must be the product of torsion groups (because they can't contain a Z term, and will obviously all be finitely generated still).
  9. Sep 30, 2011 #8
    The groups [\tex]Z108 , Z4√óZ27 [tex] are isomorphic

    Yes, but [\tex] Z108 [tex] has more music and less commercials!

    Always thought the [\tex] Z100's [tex] sound like radio stations.
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