Fundamental Theorem of Finitely Generated Abelian Groups and its application

Abelian Group problems and solutions

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 $n$.

 
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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:

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

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


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