abelian groups of order $120$ AbelianGroupsOfOrder120 2003-08-26 10:58:52 2004-03-26 10:50:34 Example fundamental theorem of finitely generated abelian groups % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} \newcommand{\Rats}{\mathbb{Q}} Here we present an application of the fundamental theorem of finitely generated abelian groups. {\bf Example (Abelian groups of order $120$)}: Let $G$ be an abelian group of order $n=120$. Since the group is finite it is obviously finitely generated, so we can apply the theorem. There exist $n_1,n_2,\ldots,n_s$ with $$G\cong \Ints/n_1\Ints\oplus\Ints/n_2\Ints\oplus\ldots\oplus\Ints/n_s\Ints$$ $$\forall i, n_i\geq 2;\quad n_{i+1}\mid n_i\ \text{for } 1\leq i\leq s-1$$ Notice that in the case of a finite group, $r$, as in the statement of the theorem, must be equal to $0$. We have $$n=120=2^3\cdot3\cdot5=\prod_{i=1}^s n_i=n_1\cdot n_2\cdot \ldots \cdot n_s$$ and by the divisibility properties of $n_i$ we must have that every prime divisor of $n$ must divide $n_1$. Thus the possibilities for $n_1$ are the following $$2\cdot 3\cdot 5,\quad 2^2\cdot 3 \cdot 5,\quad 2^3\cdot 3\cdot 5$$ If $n_1=2^3\cdot 3\cdot 5=120$ then $s=1$. In the case that $n_1=2^2\cdot 3 \cdot 5$ then $n_2=2$ and $s=2$. It remains to analyze the case $n_1=2\cdot 3 \cdot 5$. Now the only possibility for $n_2$ is $2$ and $n_3=2$ as well. Hence if $G$ is an abelian group of order $120$ it must be ({\bf up to isomorphism}) one of the following: $$\Ints/120\Ints,\quad \Ints/60\Ints\oplus \Ints/2\Ints,\quad \Ints/30\Ints\oplus\Ints/2\Ints\oplus\Ints/2\Ints$$ Also notice that they are all non-isomorphic. This is because $$\Ints/(n\cdot m)\Ints \cong \Ints/n\Ints\oplus \Ints/m\Ints \Leftrightarrow \operatorname{gcd}(n,m)=1$$ which is due to the Chinese Remainder theorem.