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Challenge 25: Finite Abelian Groups

  1. Jan 3, 2015 #1
    What is the smallest positive integer n such that there are exactly 3 nonisomorphic Abelian group of order n
  2. jcsd
  3. Jan 3, 2015 #2


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    Its 8!
  4. Jan 3, 2015 #3
    Show your work! :)
    Last edited: Jan 4, 2015
  5. Jan 3, 2015 #4


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    I thought its legitimate to use our searching skills!:D
  6. Jan 4, 2015 #5


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    n cannot be square-free (it needs factors that are multiples of each other), otherwise you don't get multiple non-isomorphic groups. The first two numbers with that property are 4 (leading to 2 different groups, corresponding to "4" and "2x2") and 8 ("8", "4x2", "2x2x2"). Therefore, 8 is the smallest n.
  7. Jan 4, 2015 #6
    Oh come on! :H:nb):oldeek: If that's not a big spoiler, I don't know what is.:oldeyes:
  8. Jan 4, 2015 #7
    Haha, I misread the challenge as asking for 3 non-Abelian groups, so - also using searching skills - I came to a different answer.
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