Question

For which positive integers $\mathrm{n}$is there exactly one abelian group of order $\mathrm{n}$(up to isomorphism)?

Short answer

The only abelian group of order $\mathrm{n}$ is ${\mathrm{\mathbb{Z}}}_{\mathrm{n}}$.

Step-by-step solution

Fundamental Theorem of Finite Abelian Group and Theorem 9.9

Fundamental Theorem of Finite Abelian Group

Every finite abelian group $\mathrm{G}$ is the direct sum of cyclic groups, each of prime power orders.

Theorem 9.9

If $n={p_1}^{n_1}{p_2}^{n_2}...{p_t}^{n_t}$ with $p_1,...p_t$ distinct primes then, ${\mathrm{\mathbb{Z}}}_n\cong {\mathrm{\mathbb{Z}}}_{{p_1}^{n_1}}\oplus ...\oplus {\mathrm{\mathbb{Z}}}_{{p_1}^{n_1}}$.

Abelian group of order n

Let $\mathrm{n}$ be the number divisible by a square of a prime number $\mathrm{p}$.

Then, there exists at least two abelian groups of order ${\mathrm{\mathbb{Z}}}_{\mathrm{p}}\oplus {\mathrm{\mathbb{Z}}}_{\mathrm{p}}\oplus {\mathrm{\mathbb{Z}}}_{\frac{\mathrm{n}}{{\mathrm{p}}^2}}$ and ${\mathrm{\mathbb{Z}}}_{{\mathrm{p}}^2}\oplus {\mathrm{\mathbb{Z}}}_{\frac{\mathrm{n}}{{\mathrm{p}}^2}}$.

If $\mathrm{n}$ is square free then, no square of a prime divides n, which implies $\mathrm{n}=\prod _{\mathrm{i}=1}^{\mathrm{k}}{\mathrm{p}}_{\mathrm{i}}$ for distinct prime numbers $\mathrm{p}$.

Thus, by Fundamental Theorem of Finite Abelian Group and Theorem 9.9, the only abelian group of order $\mathrm{n}$ is ${\mathrm{\mathbb{Z}}}_{\mathrm{n}}$.

Therefore, the only abelian group of order $\mathrm{n}$ is ${\mathrm{\mathbb{Z}}}_{\mathrm{n}}$.