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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?

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$\mathbb{Q}$... – Tyler Nov 19 '13 at 22:29

As Nicky Hekster's answer shows, the rational numbers $\mathbb{Q}$ under addition provide an example of a non-finitely generated abelian group. The classification of non-finitely generated abelian groups is an open problem. See Status of the classification of non-finitely generated abelian groups..

(A different way of seeing why $(\mathbb{Q}, +)$ is not finitely generated is to note that if $\{p_1/q_1, \ldots, p_n/q_n\}$ is a finite set of rational numbers then the additive subgroup it generates is contained in the subgroup generated by $1/q$ where $q$ is the least common multiple of the $q_i$ and hence cannot be the whole of $\mathbb{Q}$.)

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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.

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