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My question is:

For Which natural numbers $n$, a finite group $G$ of order $n$ is an abelian group?

Obviouslyو for $n≤4$ and when $n$ is a prime number, we have $G$ is abelian. Can we consider any other restrictions or conditions for $n$ to have the above statement or the group itself should have certain structure as well? Thanks.

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See math.stackexchange.com/questions/67407/… –  lhf Jun 10 '12 at 14:21
    
@lhf: Thank you for this great article. Thanks. –  B. S. Jun 10 '12 at 14:32
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It's very nice that this natural question and others along that line have a beautiful arithmetical answer. –  lhf Jun 10 '12 at 14:37
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1 Answer 1

up vote 10 down vote accepted

Every group of order $n$ is abelian iff $n$ is a cubefree nilpotent number.

We say that $n$ is a nilpotent number if when we factor $n = p_1^{a_1} \cdots p_r^{a_r}$ we have $p_i^k \not \equiv 1 \bmod{p_j}$ for all $1 \leq k \leq a_i$.

(adapted from an answer by Pete Clark.)

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Thanks for above link. After these days, this answer has been fresh yet. You know; I found a problem which was solved by J.D.Dixon in his book (Problems in Group Theory) citing: A group of order $n$ which $(n,\phi(n))=1$ is abelian. Is this the same as Pete Clark pointed in that link? Thanks and sorry for backing on this problem again. $\phi$ is Euler's function. –  B. S. Jun 23 '12 at 15:59
    
Yes, the $(n,\phi(n))=1$ criterion is the same as cubefree nilpotent. –  lhf Jun 23 '12 at 18:54
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