---- > [!definition] Definition. ([[free abelian group]]) > The **free abelian group** on a set $A$ is the '[[universal property|universal]] [[abelian group|abelian]] [[group]]' $F^{\text{ab}}(A)$ generated by the set $A$, in the sense that it satisfies the following [[universal property]]: There is a set-function $j: A \to F^{\text{ab}}(A)$ such that any function $f:A \to G$, $G$ an [[abelian group]], factors through $F^{\text{ab}}(A)$ as $f=\varphi \circ j$ for unique [[group homomorphism]] $\varphi$. In other words, the diagram > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADEA9YAHR5xgAPHMDoAjAL4SAFAEEAlCAml0mXPkIoAjOSq1GLNgHElKkBmx4CRMlr31mrRCFlK9MKAHN4RUADMAJwgAWyQAJmocCCQyfUc2ACtTfyDQxFiopB04w2c+ISw4HDgAAgBCEr56ALQACyxkkECQ8MjoxGyHXKaQagZxGAYABTUrTRAArE9anDcJIA > \begin{tikzcd} > F^{\text{ab}}(A) \arrow[r, "\exists ! \varphi"] & G \\ > A \arrow[u, "j"] \arrow[ru, "f"'] & > \end{tikzcd} > \end{document} > ``` > > commutes. [[terminal objects are unique up to a unique isomorphism|This defines]] $F^{\text{ab}}(A)$ [[isomorphism|up to]] [[group isomorphism|isomorphism]], if it exists. > > Indeed, it always exists: if $A=\{ a_{1},\dots,a_{n} \}$ is finite, then $F^{\text{ab}}(A)$ is [[group isomorphism|isomorphic]] to the [[direct sum of abelian groups|direct sum]] $\mathbb{Z}^{\oplus n}=\overbrace{\mathbb{Z} \oplus \dots \oplus \mathbb{Z}}^{n \text{ times}},$ > where the function $j: A=\{ a_{1},\dots,a_{n} \} \to \mathbb{Z}^{\oplus n}$ is given by $j(a_{i})=(0,\dots,0, \overbrace{1}^{i^{th}\text{ place}}, 0, \dots,0 ) \in \mathbb{Z}^{\oplus n}.$ > In general, $F^{\text{ab}}(A)$ is [[isomorphism|isomorphic]] to $\mathbb{Z}^{\oplus A}= \{ \alpha: A\to \mathbb{Z} : \alpha(a) \neq e_{H} \text{ for only finitely many elements } a \in A \}.$ ^definition > [!justification] Justification. > > First suppose $A$ is finite. Define the function $j: A=\{ a_{1},\dots,a_{n} \} \to \mathbb{Z}^{\oplus n}$ to be given by $j(a_{i})=(0,\dots,0, \overbrace{1}^{i^{th}\text{ place}}, 0, \dots,0 ) \in \mathbb{Z}^{\oplus n}.$ > Note that every element in $\mathbb{Z}^{\oplus n}$ may be written uniquely in the form $\sum_{i=1}^{n}m_{i}j(a_{i}).$ > Now let $f:A \to G$ be any function from $A$ to [[abelian group]] $G$. If there does exist $\varphi$ making the diagram > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \usepackage{amsfonts} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADEA9YAHR5xgAPHMDoAjAL4SAFAEEAlCAml0mXPkIoAjOSq1GLNgHElKkBmx4CRMlr31mrRCFlK9MKAHN4RUADMAJwgAWyQAJmocCCQyfUc2ACtTfyDQxFiopB04w2c+ISw4HDgAAgBCEr56ALQACyxkkECQ8MjoxGyHXKaQagZxGAYABTUrTRAArE9anDcJIA > \begin{tikzcd} > \mathbb{Z}^{\oplus n} \arrow[r, "\varphi"] & G \\ > A \arrow[u, "j"] \arrow[ru, "f"'] & > \end{tikzcd} > \end{document} > ``` > commute, then commutativity enforces $\begin{align} > f(a_{i})=&\varphi \circ j(a_{i}) \\ > = & \varphi(0, \dots, 0,1,0,\dots,0) > \end{align}$ meaning that $\varphis value on the 'basis element' $(0,\dots,0, \overbrace{1}^{i^{th} \text{ place}},0,\dots,0)$ of $\mathbb{Z}^{\oplus n}$ is determined as $i^{th} \text{ basis element} \mapsto f(a_{i})$. But since $\varphi$ is a [[group homomorphism]], this actually enforces its behavior entirely: given an arbitrary $\boldsymbol k=\sum_{i=1}^{n} m_{i}j(a_{i}) \in \mathbb{Z}^{\oplus n}$, we have to have $\begin{align} > \varphi(\boldsymbol k) = & \varphi\left( \sum_{i=1}^{n} m_{i}j(a_{i}) \right) \\ > = & \sum_{i=1}^{n} \varphi(\overbrace{j(a_{i}) + \dots + j(a_{i})}^{m_{i} \text{ times}}) \\ > = & \sum_{i=1}^{n} m_{i} \varphi(j(a_{i})) \\ > =& \sum_{i=1}^{n} m_{i} f(a_{i}). > \end{align}$ > Thus, should $\varphi$ exist, it is unique. But *does* it exist? We just have to check that the function we've defined is a [[group homomorphism]]. This is where commutativity of $G$ enters: $\begin{align} > \varphi\left( \sum_{i=1}^{n}m_{i}' j(a_{i}) \right) + \varphi\left( \sum_{i=1}^{n} m_{i}''j(a_{i}) \right) = & \sum_{i=1}^{n} m_{i}'f(a_{i} ) + \sum_{i=1}^{n} m_{i}''f(a_{i}) \\ > \textcolor{Thistle}{=} & \sum_{i=1}^{n} (m_{i}' + m_{i}'')f(a_{i}) \\ > = & \varphi\left( \sum_{i=1}^{n} (m_{i}' + m_{i}'')j(a_{i}) \right) \\ > = & \varphi\left( \sum_{i=1}^{n} m_{i}' j(a_{i}) + \sum_{i=1}^{n} m_{i}j(a_{i}) + \sum_{i=1}^{n} m_{i}''j(a_{i}) \right). > \end{align}$ > For the general case, $j:A \to \mathbb{Z}^{\oplus A}$ is defined so that $j(a)=(A \xrightarrow{j_{a}}\mathbb{Z})$, where the function $j_{a} \in \mathbb{Z}^{\oplus A}$ is given by the indicator $j_{a}(x):= \begin{cases} 0 & x \neq a; \\ 1 & x=a. \end{cases}$ The key observation is that an arbitrary element $\boldsymbol k$ of $\mathbb{Z}^{\oplus A}$ may be uniquely written as a *finite* sum $\boldsymbol k=\sum_{ a\in A} m_{a} j(a), m_{a} \neq 0 \text{ for only finitely many }a.$ Now all the arguments provided for the finite case carry through. ^justification ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```