AbelianGroupIsDivisibleIfAndOnlyIfItIsAnInjectiveObject 2009-02-06 15:18:22 2009-05-31 08:58:10 Theorem divisible group % 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{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 \textbf{Proposition.} Abelian group $A$ is divisible if and only if $A$ is an injective object in the category of abelian groups. \textit{Proof.} ,,$\Leftarrow$'' Assume that $A$ is not divisible, i.e. there exists $a\in A$ and $n\in\mathbb{N}$ such that the equation $nx=a$ has no solution in $A$. Let $B=<a>$ be a cyclic subgroup generated by $a$ and $i:B\to A$ the canonical inclusion. Now there are two possibilities: either $B$ is finite or infinite. If $B$ is infinite, then let $H=\mathbb{Z}$ and let $f:B\to H$ be defined on generator by $f(a)=n$. Now $A$ is injective, thus there exists $h:H\to A$ such that $h\circ f=i$. Thus $$n\cdot h(1)=h(1)+\cdots +h(1)=h(1+\cdots +1)=h(n)=h(f(a))=i(a)=a.$$ Contradiction with definition of $n\in\mathbb{N}$ and $a\in A$. If $B$ is finite, then let $k=|B|$ (note that $n$ does not divide $k$) and let $H=\mathbb{Z}_{n\cdot k}$. Furtheremore define $f:B\to H$ on generator by $f(a)=n$ (note that in this case $f$ is a well defined homomorphism). Again injectivity of $A$ implies existence of $h:H\to A$ such that $h\circ f=i$. Similarly we get contradiction: $$n\cdot h(1)=h(1)+\cdots +h(1)=h(1+\cdots +1)=h(n)=h(f(a))=i(a)=a.$$ This completes first implication. ,,$\Rightarrow$'' This implication is proven \PMlinkname{here}{ExampleOfInjectiveModule}. $\square$ \textbf{Remark.} It is clear that in the category of abelian groups $\mathcal{AB}$, a group $A$ is projective if and only if $A$ is free. This is since $\mathcal{AB}$ is equivalent to the category of $\mathbb{Z}$-modules and projective modules are direct summands of free modules. Since $\mathbb{Z}$ is a principal ideal domain, then every submodule of a free module is free, thus projective $\mathbb{Z}$-modules are free.