The Structure Theorem: Every finitely generated abelian group is isomorphic to a direct product of cyclic groups $C_{d_0}\times C_{d_1}\times \ldots \times C_{d_k} \times L$ such that $d_i | d_{i+1} \forall \space 0\le i\le {k-1}$ and $L$ is a free abelien groups (i.e $\mathbb{Z}^r$ for some $r$). One method to prove this fact is using Smith Normal Form. What is an example of a abelian groups that is NOT finitely generated? What can we say about about the isomorphism classes of infinitly generated abelien groups?
[Math] Abelian Group Not Finitely Generated
group-theorysoft-question
Related Solutions
The fundamental theorem of finitely generated abelian groups does not say that you can split up $360$ in any way you want as a product of primes and obtain an isomorphism. It says that for any abelian group of order $360$, there is a way to write $360$ as a product of prime powers $360=p_1^{r_1}\cdot\cdots \cdot p_n^{r_n}$ such that $G\cong \mathbb Z_{p_1^{r_1}}\times Z_{p_n^{r_n}}$.
See the Wikipedia article on finitely-generated abelian groups for more information. We have
Theorem A. Any finite abelian group $G$ admits a unique invariant factor decomposition.
Theorem B. Any finite abelian group $G$ admits a unique primary decomposition.
These decompositions are different, but logically Thm A $\Leftrightarrow$ Thm B: that is, once you prove one, the other follows from Sun-Ze (aka CRT), ${\bf Z}/nm\cong {\bf Z}/n\times{\bf Z}/m$ when $(n,m)=1$. Further one can compute one decomposition from the other (+ vice-versa). Let's discuss how.
Suppose we have the invariant factor decomposition of $G$ from Thm A as follows:
$$G\cong \frac{\bf Z}{n_1}\times\frac{\bf Z}{n_2}\times\cdots\times\frac{\bf Z}{n_l}.$$
Then each $n_i$ factors as $\prod_pp^{e(i,p)}$ for primes $p$ and exponents $e(i,p)$. By SZ then
$$G\cong\prod_{i=1}^l\frac{\bf Z}{n_l}\cong\prod_{i=1}^l\prod_p\frac{\bf Z}{p^{e(i,p)}}\cong\prod_p\left[\prod_{i=1}^l\frac{\bf Z}{p^{e(i,p)}}\right].$$
Notice the factors $\prod_{i=1}^l{\bf Z}/p^{e(i,p)}$ are $p$-groups; this is the primary decomposition, which is the same as the Sylow decomposition (as a direct product of Sylow $p$-subgroups, available for all nilpotent groups) when our group is abelian. It is probably best to think of Sylow theory as the inevitable noncommutative generalization of arithmetic in the theory of finitely-generated abelian groups.
Our original hypothesis that we had the invariant factor decomposition means that $n_i\mid n_{i+1}$, so we must have $e(i,p)\le e(i+1,p)$ for each $p$. Also, some of the $e$'s may be $0$. Thus if we begin with a primary decomposition as regarded in Thm B, we may order the prime exponents in an increasing order, and fill in extra $0$s as needed so that each prime has the same number of exponents, then form the $n_i$ out of the prime powers created from these exponents.
Here's an example. From a primary decomposition we obtain
$$\begin{array}{llccccc} G & \cong & \left(\frac{\bf Z}{2} \times \frac{\bf Z}{2}\times\frac{\bf Z}{8}\right) & \times & \left(\frac{\bf Z}{3}\times\frac{\bf Z}{27}\right) & \times & \left(\frac{\bf Z}{5}\right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}} \times \color{Blue}{\frac{\bf Z}{2}} \times \color{Green}{\frac{\bf Z}{8}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{3}} \times \color{Green}{\frac{\bf Z}{27}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{1}} \times \color{Green}{\frac{\bf Z}{5}} \right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}\times\frac{\bf Z}{1}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Blue}{\frac{\bf Z}{2}\times\frac{\bf Z}{3}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Green}{\frac{\bf Z}{8}\times\frac{\bf Z}{27}\times\frac{\bf Z}{5}} \right) \\ & \cong & \frac{\bf Z}{2\times1\times1} & \times & \frac{\bf Z}{2\times3\times1} & \times & \frac{\bf Z}{8\times27\times5} \\[7pt] & \cong & {\bf Z}/2 & \times & {\bf Z}/6 & \times & {\bf Z}/1080 \end{array}$$
as our invariant factor decomposition, with $2\mid6\mid1080$. And to see how to go in reverse with this same example, just read from bottom to top.
In conclusion, then, there are two standard fundamental representations of a finite abelian group, the invariant-factor decomposition and the primary decomposition. Theorems A and B state their existence and uniqueness; these are roughly logically equivalent using the remainder theorem, and it is relatively easy to move between the two decompositions using prime factorization. The IF representation invokes linear algebra, the primary decomposition invokes group theory (it is positioned within Sylow theory), and they are both positioned within commutative algebra.
Best Answer
Assume that $(\mathbb Q,+)$ is finitely generated. Then, by the structure theorem there exist $m,n\ge 0$ and $d_1\mid\cdots\mid d_m$ with $d_i>1$ such that $\mathbb Q\simeq \mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_m\mathbb Z\oplus \mathbb Z^n$. If $m\ge 1$, then there exists $x\in\mathbb Q$, $x\neq 0$, such that $d_1x=0$, a contradiction. Thus we get $m=0$. Then $\mathbb Q\simeq \mathbb Z^n$. If $n\ge 2,$ then there exist $x_1,x_2\in\mathbb Q$ which are linearly independent over $\mathbb Z$. But $x_1=a_1/b_1$ and $x_2=a_2/b_2$ give $(b_1a_2)x_1+(-b_2a_1)x_2=0$, a contradiction. So we must have $n=1$, that is, $\mathbb Q$ is cyclic. Assume that it is generated by $a/b$ with $b\ge 1$. Then $\frac{1}{b+1}$ can not be written as $\frac{ka}{b}$, and again a contradiction.