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

**crushingyen** · February 25th 2010, 04:20 PM

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

**Drexel28** · February 25th 2010, 04:29 PM

Originally Posted by crushingyen

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, $G\simeq\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z }_{p_k^{\alpha}}\oplus\cdots\oplus\mathbb{Z}_{p_m} \simeq\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z} _{p_k}\oplus\mathbb{Z}_{p_k^{\alpha-1}}\oplus\cdots\oplus\mathbb{Z}_{p_m}$

**crushingyen** · February 25th 2010, 08:48 PM

You are amazing. Thanks!

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