Every finitely generated abelian group $A$ is a direct sum of its cyclic subgroups, i.e. $$A=C_{m_1}\oplus C_{m_2}\oplus\ldots\oplus C_{m_k}\oplus \mathbb{Z}\oplus\ldots\oplus\mathbb{Z},$$ where $1<m_1 \mid m_2 \mid \ldots \mid m_k$ The numbers $m_i$ are uniquely determined as well as the number of $\mathbb{Z}$ s, which is the rank of an abelian group.

Proof. Let $G$ be an abelian group with $n$ generators. Then for a free group $F_n$ $G$ is isomorphic to the quotient group $F_n/A$ Now $F_n$ and $A$ contain a basis $f_1,\ldots,f_n$ and $a_1,\ldots,a_k$ satisfying $a_i=m_if_i$ for all $1\leq i\leq k$ As $G\cong F_n/A$ it suffices to show that $F_n/A$ is a direct sum of its cyclic subgroups $\langle f_1+A\rangle$

It is clear that $F_n/A$ is generated by its subgroups $\langle f_i+A\rangle$ Assume that the zero element of $F_n/A$ can be written as a form $A=l_1f_1+\ldots+l_nf_n+A$ It follows that $l_1f_1+\ldots+l_nf_n=a\in A$ As we write $a$ as a linear combination of that basis and using $a_i=m_if_i$ we get the equations $$l_1f_1+\ldots+l_nf_n=s_1a_1+\ldots s_ka_k=s_1m_1f_1+\ldots +s_km_kf_k.$$ As every element can be represented uniquely as a linear combination of its free generators $f_1$ we have $l_i=s_im_i$ for every $1\leq i\leq k$ and $l_j=0$ for every $k<j\leq n$

This means that every element $l_if_i$ belongs to $A$ so $l_if_i+A=A$ Therefore the zero element has a unique representation as a sum of the elements of the subgroup $\langle f_i+\!A\rangle$

Bibliography

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P. PAAJANEN: Ryhmäteoria. Lecture notes, Helsinki university, Finland (fall 2008)