Here we present an application of the fundamental theorem of finitely generated abelian groups.

Example (Abelian groups of order $120$ ):

Let $G$ be an abelian group of order $n=120$ . Since the group is finite it is obviously finitely generated, so we can apply the theorem. There exist $n_1,n_2,\ldots,n_s$ with $$G\cong \Ints/n_1\Ints\oplus\Ints/n_2\Ints\oplus\ldots\oplus\Ints/n_s\Ints$$ $$\forall i, n_i\geq 2;\quad n_{i+1}\mid n_i\ \text{for } 1\leq i\leq s-1$$ Notice that in the case of a finite group, $r$ , as in the statement of the theorem, must be equal to $0$ . We have $$n=120=2^3\cdot3\cdot5=\prod_{i=1}^s n_i=n_1\cdot n_2\cdot \ldots \cdot n_s$$ and by the divisibility properties of $n_i$ we must have that every prime divisor of $n$ must divide $n_1$ . Thus the possibilities for $n_1$ are the following $$2\cdot 3\cdot 5,\quad 2^2\cdot 3 \cdot 5,\quad 2^3\cdot 3\cdot 5$$ If $n_1=2^3\cdot 3\cdot 5=120$ then $s=1$ . In the case that $n_1=2^2\cdot 3 \cdot 5$ then $n_2=2$ and $s=2$ . It remains to analyze the case $n_1=2\cdot 3 \cdot 5$ . Now the only possibility for $n_2$ is $2$ and $n_3=2$ as well.

Hence if $G$ is an abelian group of order $120$ it must be (up to isomorphism) one of the following: $$\Ints/120\Ints,\quad \Ints/60\Ints\oplus \Ints/2\Ints,\quad \Ints/30\Ints\oplus\Ints/2\Ints\oplus\Ints/2\Ints$$ Also notice that they are all non-isomorphic. This is because $$\Ints/(n\cdot m)\Ints \cong \Ints/n\Ints\oplus \Ints/m\Ints \Leftrightarrow \operatorname{gcd}(n,m)=1$$ which is due to the Chinese Remainder theorem.