Originally Posted by

**crushingyen** For which $\displaystyle n$ is there exactly one abelian group of order $\displaystyle n$?

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

Thus, for $\displaystyle n$ a product of distinct primes, there is one abelian group of order $\displaystyle n$.

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

Thanks,

crushingyen