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Math Help - For which n is there exactly one abelian group of order n?

  1. #1
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    For which n is there exactly one abelian group of order n?

    For which n is there exactly one abelian group of order n?

    I know that the answer is the product of distinct primes, but I need to say this in a formal proof.
    I started out the proof doing this:

    Suppose there is a group G such that |G| = n = p_1^{n_1}p_2^{n_2} \cdot \cdot \cdot p_k^{n_k}. The elementary divisors are uniquely determined by the order of the group if and only if p_1, p_2, ..., p_k are distinct primes. So then G can be written uniquely in one way - the direct sum of its cyclic groups.
    Thus, for n a product of distinct primes, there is one abelian group of order n.

    Is there something I can do to make this answer stronger?

    Thanks,
    crushingyen
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by crushingyen View Post
    For which n is there exactly one abelian group of order n?

    I know that the answer is the product of distinct primes, but I need to say this in a formal proof.
    I started out the proof doing this:

    Suppose there is a group G such that |G| = n = p_1^{n_1}p_2^{n_2} \cdot \cdot \cdot p_k^{n_k}. The elementary divisors are uniquely determined by the order of the group if and only if p_1, p_2, ..., p_k are distinct primes. So then G can be written uniquely in one way - the direct sum of its cyclic groups.
    Thus, for n a product of distinct primes, there is one abelian group of order n.

    Is there something I can do to make this answer stronger?

    Thanks,
    crushingyen
    You know the fundamental theorem of finitely generated abelian groups? Then, if G is abelian and |n|=p_1\cdots p_m then G\simeq \bigoplus_{k=1}^{m}\mathbb{Z}_{p_k}. And it is clear that this is the only decomposition. Thus, every group with whose order is the product of distinct primes surely satisfies this. Otherwise, suppose that |n|=p_1\cdots p_k^{\alpha}\cdots p_m where \alpha>1. Then, G\simeq\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z  }_{p_k^{\alpha}}\oplus\cdots\oplus\mathbb{Z}_{p_m}  \simeq\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z}  _{p_k}\oplus\mathbb{Z}_{p_k^{\alpha-1}}\oplus\cdots\oplus\mathbb{Z}_{p_m}
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    You are amazing. Thanks!
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