Why are not ALL abelian groups $G\simeq \mathbb Z_{n_1} \times \mathbb Z_{n_2} \times \cdots\times \mathbb Z_{n_s}$ where $|G| = n_1 n_2 \cdots n_s$?

Question

I'm in the middle of reading this text, but something bugs me. That is:

Why are not ALL abelian groups $G\simeq \mathbb Z_{n_1} \times \mathbb Z_{n_2} \times \cdots\times \mathbb Z_{n_s}$ where $|G| = n_1 n_2 \cdots n_s$?

For example for $|G| = 180$, the abelian groups are: $Z_{180}$, $Z_{90} \times Z_2$, $Z_{60} \times Z_3$ and $Z_{30} \times Z_6$.

Why is not for example $Z_{18} \times Z_{10}$?

Here is the text:

Answer 1

They're just not. Groups of the form $\mathbb Z_n$ are cyclic: if you keep adding the class of $1$ to itself, you get everything in the group. Example: $\mathbb Z_4$ the elements are:

$$1$$

$$1+1$$

$$1+1+1$$

$$1+1+1+1$$

A group like $\mathbb Z_2 \times \mathbb Z_2$ does not have an element like that. The elements are $(0,0);(1,0);(0,1);(1,1)$ and if you continually add any of these elements to itself, you will not get all the elements of the group. So $\mathbb Z_2 \times \mathbb Z_2$ cannot be isomorphic to a group like $\mathbb Z_n$.

Answer 2

Consider the set of integers $\mathbb{Z}$ as an additive group. It's abelian, but it simply is not of the form $Z_{n_1} \times Z_{n_2} \times ... \times Z_{n_s}$. (If it were of that form, it would be a finite group.)

Consider the set of rational numbers $\mathbb{Q}$ as an additive group. It's abelian, but it simply is not of the form $\mathbb{Z}^r\times Z_{n_1} \times Z_{n_2} \times ... \times Z_{n_s}$. (If it were of that form, then either $r=0$ and $\mathbb{Q}$ would be finite, or $r\geq1$ and you couldn't divide certain elements like $(1,0,0,\ldots,0)$ in half.)