proof of fundamental theorem of finitely generated abelian groups

Every finitely generated abelian group $A$ is a direct sum of its cyclic subgroups, i.e.

$$A=C_{m_1}\oplus C_{m_2}\oplus \mathrm{\dots }\oplus C_{m_k}\oplus \mathbb{Z}\oplus \mathrm{\dots }\oplus \mathbb{Z},$$

where  . 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 $f_1,\mathrm{\dots },f_n$ and $a_1,\mathrm{\dots },a_k$ satisfying $a_i=m_i{f}_i$ for all $1\le i\le k$. As $G\cong F_n/A$, it suffices to show that $F_n/A$ is a direct sum of its cyclic subgroups $⟨f_1+A⟩$.

It is clear that $F_n/A$ is generated by its subgroups $⟨f_i+A⟩$. Assume that the zero element of $F_n/A$ can be written as a form $A=l_1{f}_1+\mathrm{\dots }+l_n{f}_n+A$. It follows that $l_1{f}_1+\mathrm{\dots }+l_n{f}_n=a\in A$. As we write $a$ as a linear combination of that basis and using $a_i=m_i{f}_i$ we get the equations

$$l_1{f}_1+\mathrm{\dots }+l_n{f}_n=s_1{a}_1+\mathrm{\dots }{s}_k{a}_k=s_1{m}_1{f}_1+\mathrm{\dots }+s_k{m}_k{f}_k.$$

As every element can be represented uniquely as a linear combination of its free generators $f_1$, we have $l_i=s_i{m}_i$ for every $1\le i\le k$ and $l_j=0$ for every .

This means that every element $l_i{f}_i$ belongs to $A$, so  $l_i{f}_i+A=A$. Therefore the zero element has a unique representation as a sum of the elements of the subgroup $⟨f_i+A⟩$.