# Fundamental Theorem of Finitely Generated Abelian Groups and its application

## Problem 420

In this post, we study the **Fundamental Theorem of Finitely Generated Abelian Groups**, and as an application we solve the following problem.

**Problem**.

Let $G$ be a finite abelian group of order $n$.

If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$.

Contents

## Fundamental Theorem of Finitely Generated Abelian Groups

Before stating the fundamental theorem for finitely generated abelian groups, we define several terminologies and notations.

### Definitions / notations

- We say that a group $G$ is finitely generated if there is a finite subset $S$ of $G$ such that $G$ is generated by $S$, that is, $G=\langle S \rangle$.
- For each positive integer $r$, let

\[\Z^r=\underbrace{\Z\times \Z \times \cdots \times \Z}_{\text{$r$ times}}\] be the direct product of $r$ copies of $\Z$. Here we set $\Z^0=1$ to be the trivial group. - The group $\Z^r$ is called the free abelian group of rank $r$.
- For each positive integer $n$, let $\Z_n=\Zmod{n}$ be the cyclic group of order $n$.

### Theorem (Fundamental Theorem of Finitely Generated Abelian Groups)

**Theorem**. Let $G$ be a finitely generated abelian group. Then it decomposes as follows:

\[G\cong \Z^r\times Z_{n_1}\times \Z_{n_2}\times \cdots \times \Z_{n_s}, \tag{*}\] for some integers $r, n_1, n_2, \dots, n_s$ satisfying the following conditions:

- $r\geq 0$ and $n_i \geq 2$ for all $i$, and
- $n_{i+1}|n_i$ for $1 \leq i \leq s-1$.

The decomposition of $G$ satisfying these conditions is unique.

- The integer $r$ in the decomposition (*) is called the
**free rank**or**Betti number**of $G$. - The integers $n_1, n_2, \dots, n_s$ are called the
**invariant factors**of $G$. - The decomposition (*) is called the
**invariant factor decomposition**of $G$.

## Problem

Let $G$ be a finite abelian group of order $n$.

If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$.

## Proof.

Since $G$ is a finite abelian group, it is in particular a finitely generated abelian group.

(We can take $G$ itself for a finite set of generators of $G$.)

Then by the fundamental theorem of finitely generated abelian groups, we have the invariant factor decomposition

\[G\cong \Z^r\times Z_{n_1}\times \Z_{n_2}\times \cdots \times \Z_{n_s}\]
satisfying

- $r\geq 0$ and $n_i \geq 2$ for all $i$, and
- $n_{i+1}|n_i$ for $1 \leq i \leq s-1$.

Since $G$ is a finite group, the rank $r$ must be $0$. Thus we have an isomorphism

\[G\cong Z_{n_1}\times \Z_{n_2}\times \cdots \times \Z_{n_s}.\]
Comparing the order, we have

\[n=n_1 n_2\cdots n_s.\]

Let $p$ be a prime factor of $n$. Then $p$ divides some $n_i$.

If $i>1$, then it follows from condition (2) that $p$ divides $n_1$ as well.

Thus $p^2$ divides $n$. Since $n$ is a square-free integer, this is a contradiction.

It follows that any prime factor of $n$ divides only $n_1$.

Therefore we obtain $n=n_1$ and $s=1$. So the invariant factor decomposition of $G$ is

\[G\cong Z_n.\]
Hence $G$ is isomorphic to the cyclic group $\Z_n$ of order $n$.

## Reference

Abstract Algebra by Dummit and Foote (third edition) Section 5.2.

Add to solve later

Sponsored Links