# proof of fundamental theorem of finitely generated abelian groups

 $A=C_{m_{1}}\oplus C_{m_{2}}\oplus\ldots\oplus C_{m_{k}}\oplus\mathbb{Z}\oplus% \ldots\oplus\mathbb{Z},$

where  $1. 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 $a_{i}=m_{i}f_{i}$ 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 $A=l_{1}f_{1}+\ldots+l_{n}f_{n}+A$. It follows that $l_{1}f_{1}+\ldots+l_{n}f_{n}=a\in A$. As we write $a$ as a linear combination  of that basis and using $a_{i}=m_{i}f_{i}$ we get the equations

 $l_{1}f_{1}+\ldots+l_{n}f_{n}=s_{1}a_{1}+\ldots s_{k}a_{k}=s_{1}m_{1}f_{1}+% \ldots+s_{k}m_{k}f_{k}.$

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 $1\leq i\leq k$ and $l_{j}=0$ for every $k.

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 $\langle f_{i}+\!A\rangle$.

## References

• 1 P. Paajanen: Ryhmäteoria.  Lecture notes, Helsinki university, Finland (fall 2008)
Title proof of fundamental theorem of finitely generated abelian groups ProofOfFundamentalTheoremOfFinitelyGeneratedAbelianGroups 2013-03-22 18:24:58 2013-03-22 18:24:58 puuhikki (9774) puuhikki (9774) 7 puuhikki (9774) Proof msc 20K25