# abelian groups of order $120$

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

 $G\cong\mathbb{Z}/n_{1}\mathbb{Z}\oplus\mathbb{Z}/n_{2}\mathbb{Z}\oplus\ldots% \oplus\mathbb{Z}/n_{s}\mathbb{Z}$
 $\forall i,n_{i}\geq 2;\quad n_{i+1}\mid n_{i}\ \text{for }1\leq i\leq s-1$

Notice that in the case of a finite group, $r$, as in the statement of the theorem, must be equal to $0$. We have

 $n=120=2^{3}\cdot 3\cdot 5=\prod_{i=1}^{s}n_{i}=n_{1}\cdot n_{2}\cdot\ldots% \cdot n_{s}$

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:

 $\mathbb{Z}/120\mathbb{Z},\quad\mathbb{Z}/60\mathbb{Z}\oplus\mathbb{Z}/2\mathbb% {Z},\quad\mathbb{Z}/30\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2% \mathbb{Z}$

Also notice that they are all non-isomorphic. This is because

 $\mathbb{Z}/(n\cdot m)\mathbb{Z}\cong\mathbb{Z}/n\mathbb{Z}\oplus\mathbb{Z}/m% \mathbb{Z}\Leftrightarrow\operatorname{gcd}(n,m)=1$

which is due to the Chinese Remainder theorem.

Title abelian groups of order $120$ AbelianGroupsOfOrder120 2013-03-22 13:54:17 2013-03-22 13:54:17 alozano (2414) alozano (2414) 5 alozano (2414) Example msc 20E34 FundamentalTheoremOfFinitelyGeneratedAbelianGroups AbelianGroup2