Challenge 25: Finite Abelian Groups

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SUMMARY

The smallest positive integer n such that there are exactly 3 nonisomorphic Abelian groups of order n is 8. This conclusion is derived from the requirement that n cannot be square-free, necessitating factors that are multiples of each other. The analysis identifies that the first two integers with this property are 4, which yields 2 nonisomorphic groups, and 8, which yields 3 nonisomorphic groups: "8", "4x2", and "2x2x2". Thus, 8 is confirmed as the correct answer.

PREREQUISITES
  • Understanding of Abelian group theory
  • Knowledge of group isomorphism
  • Familiarity with the classification of finite groups
  • Basic number theory concepts related to divisibility
NEXT STEPS
  • Study the classification theorem for finite Abelian groups
  • Explore the properties of group orders and their implications
  • Learn about the structure of nonisomorphic groups
  • Investigate examples of finite groups and their classifications
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Mathematicians, students of abstract algebra, and anyone interested in the properties and classifications of finite groups will benefit from this discussion.

Greg Bernhardt
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What is the smallest positive integer n such that there are exactly 3 nonisomorphic Abelian group of order n
 
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Shyan said:
Its 8!
Show your work! :)
 
Last edited:
Greg Bernhardt said:
Show you're work! :)
I thought its legitimate to use our searching skills!:D
 
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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Shyan said:
Its 8!

I thought its legitimate to use our searching skills
Oh come on! :H:nb):oldeek: If that's not a big spoiler, I don't know what is.:oldeyes:
 
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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