abelian groups of order $120$

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

Example (Abelian groups of order $\mathrm{120}$):

Let $G$ be an abelian group of order $n=120$. Since the group isfinite it is obviously finitely generated, so we can apply thetheorem. There exist $n_1,n_2,\mathrm{\dots },n_s$ with

$$G\cong \mathbb{Z}/n_1\mathbb{Z}\oplus \mathbb{Z}/n_2\mathbb{Z}\oplus \mathrm{\dots }\oplus \mathbb{Z}/n_s\mathbb{Z}$$

$$\forall i,n_i\ge 2;n_{i+1}\mid n_i\text{for }1\le i\le s-1$$

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

$$n=120=2^3\cdot 3\cdot 5=\prod _{i=1}^s{n}_i=n_1\cdot n_2\cdot \mathrm{\dots }\cdot n_s$$

and by the divisibility properties of $n_i$ we must have thatevery prime divisor of $n$ must divide $n_1$. Thus thepossibilities for $n_1$ are the following

$$2\cdot 3\cdot 5,2^2\cdot 3\cdot 5,2^3\cdot 3\cdot 5$$

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 toanalyze 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:

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

Also noticethat they are all non-isomorphic. This is because

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

which is due to theChinese Remainder theorem.