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What is the smallest positive integer n such that there are exactly 3 nonisomorphic Abelian group of order n
Challenge 25: Finite Abelian Groups
- #2
ShayanJ
Gold Member
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Its 8!
- #3
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Show your work! :)
Last edited:
- #4
ShayanJ
Gold Member
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I thought its legitimate to use our searching skills!:D
- #5
mfb
Mentor
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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.
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If that's not a big spoiler, I don't know what is.
- #7
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Haha, I misread the challenge as asking for 3 non-Abelian groups, so - also using searching skills - I came to a different answer.