Challenge 25: Finite Abelian Groups

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The smallest positive integer n for which there are exactly three nonisomorphic Abelian groups of order n is 8. This conclusion is based on the requirement that n cannot be square-free, necessitating factors that are multiples of each other. The groups corresponding to n=8 are "8", "4x2", and "2x2x2". The discussion also highlights a misunderstanding where one participant initially misinterpreted the challenge as seeking non-Abelian groups. Ultimately, the consensus confirms that 8 is indeed the correct answer.
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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