---- > [!definition] Definition. ([[free group]]) > The **free group** on a set $A$ is the '[[universal property|universal]] [[group]]' $F(A)$ [[generating set of a group|generated by]] the set $A$, in the sense that it satisfies the following [[universal property]]: There is a set-function $j:A \to F(A)$ such that any function $f: A \to G$, $G$ any [[group]], factors through $F(A)$ as $f=\varphi \circ j$ for unique [[group homomorphism]] $\varphi$, i.e., the diagram > > ```tikz > \usepackage{tikz-cd} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADEAKAQQEoQAvqXSZc+QigCM5KrUYs2AcUHCQGbHgJEyk2fWatEIboNkwoAc3hFQAMwBOEALZIATNRwQk0uQba2QagY6ACMYBgAFUU0JEHssCwALHBU7RxdEMhBPb2p9BSMAHUKYAA8sOBw4AAIAQmri+ns0RKxUkAdnNw8vTLz5QxAAK1MBIA > \begin{tikzcd} > F(A) \arrow[r, "\exists ! \varphi"] & G \\ > A \arrow[ru, "f"'] \arrow[u, "j"] & > \end{tikzcd} > \end{document} > ``` > > commutes. [[terminal objects are unique up to a unique isomorphism|This defines]] $F(A)$ [[isomorphism|up to]] [[group isomorphism|isomorphism]], if it exists. > > Indeed, it always exists: denoting by $R(A):W(A) \to W(A)$ the [[reduced word|reduction map]], taking $F(A)$ to be the set $\text{im }R$ of reduced words on $A$, endowed with [[binary operation]] given by *juxtaposition & reduction* $w_{1} \cdot w_{2} := R(w_{1}w_{2}),$and $j$ the 'string literalization map' $a \xmapsto{j} \text{`}a\text{'}$, satisfies the [[universal property]] above. > [!note] Remark 1. > In order to establish a suitable notion of '$F(A)$ contains $A, one might expect this definition to enforce $j$ to be [[injection|injective]]. Indeed, the [[universal property]] guarantees this: suppose $a$ and $b$ are two elements of $A$. Let $f:A \to G=C_{2}$ kill $a$ and send everything else to the generator. Then $f(a) \neq f(b)$, hence $\varphi \circ j(a) \neq \varphi \circ j(b)$, and hence $j(a) \neq j(b)$. ^1cd9e3 > [!note] Remark 2. > The free group construction is a [[covariant functor|functor]] $\mathscr{F}:\mathsf{Set} \to \mathsf{Grp}$. Indeed, given a set $A$, declare $\mathscr{F}A:=F(A)$, and given a set-function $g:A \to B$, declare $\mathscr{F}g:=\varphi$, where $\varphi$ is (uniquely) induced by the [[free group]] [[universal property]] as in the diagram below: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKchWp0mrdgCEefEBmx4CRMsXEMWbRCABiACg4BKOfyVCio9TU1SdB6Se7iYUAObwioAGYAnCAC2SABMNDgQSADMNIxYYNogUPRwABauIDaSCQA62TAAHlhwOHAABACEpbkMvmgpWKYgfoFIZCDhIZla7ABWAPpcvD7+QYii7RGI0RLdOv2yQ00jrWGT47YJbhkgjPQARjCMAAoCysIgvlhuKTiNzaNtHVM0B2BQSAC0kW0b7N6liAAvKV5lVsgBjLC+cGlLYxfaHE7mFQ6Rgwby3ZzcIA > \begin{tikzcd} > F(A) \arrow[r, "\exists ! \varphi", dashed] & F(B) \\ > A \arrow[u, "j_A"] \arrow[r, "g"'] \arrow[ru, "f := j_B \circ g", bend right] & B \arrow[u, "j_B"] > \end{tikzcd} > \end{document} > ``` > To check functoriality, note that $\mathscr{F}(\id_{A})=\id_{F(A)}$ > (the diagram above 'collapses') so that identities are preserved. For compositions, contemplate the following diagram: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKchWp0mrdgCEefEBmx4CRAExiaDFm0QgAwnP5KhRMsXFapugGIAKDgEpDCgcuHJR5zZJ0g70p14jQRUUdS8JbXY7PUDxGCgAc3giUAAzACcIAFskMhAcCCQAZiCQTJykUQKixAAWMorcxHUapABWRqzm-MKq7yjdRJAaRnoAIxhGAAVXE10MrESACxxnJv62loGrEGWRkDHJmbnQkEWVta7KxGKaPvqd3wAdZ5gADyw4HDgAQgACV4MDJoZZYAD6w2uzTq91q7Se7FeHy+P3+AKB9BBYP+4P20JKcI6NEmYCgJXylheb0+3zg6MBz2BoKwPEo3CAA > \begin{tikzcd} > F(A) \arrow[r, "\exists! \varphi_g"] \arrow[rr, "\exists ! \varphi", bend left] & F(B) \arrow[r, "\exists ! \varphi _h"] & F(C) \\ > A \arrow[u] \arrow[r, "g"'] & B \arrow[u] \arrow[r, "h"'] & C \arrow[u] > \end{tikzcd} > \end{document} > ``` > *uniqueness* of the induced homomorphism ensures that $\mathscr{F}$ preserves compositions — for $\mathscr{F}(hg)=\varphi$ is the unique map from $F(A)$ to $F(C)$ making the diagram commute, but the map $\mathscr{F}(h)\mathscr{F}(g)=\varphi_{h} \varphi_{g}$ also makes the diagram commute, so in fact these two maps must be equal. ^3e62b9 > [!intuition] > In a sense, $F(A)$ is the 'most efficient' way to construct a [[group]] containing $A$, since any other way to map $A$ to a [[group]] $G$ can be reconstructed from this one via $\varphi$. Put differently, any way to map $A$ to a [[group]] $G$ must 'uniquely factor through' $F(A)$. ^intuition ---- #### > [!justification] > First we have to (1) make precise the [[universal property]] as stated. Then we have to (2) show that the [[binary operation]] $\cdot$ turns $\im R$ into a [[group]]. Finally, we must verify (3) that $\big( j, \im R \big)$ indeed satisfies the [[universal property]]. > > **1.** Consider the [[category]] $\mathscr{F}^{A}$ whose objects are pairs $(j, G)$, where $G$ is a [[group]] and $j: A \to G$ is a set-function and whose morphisms $(j_{1}, G_{1}) \to (j_{2}, G_{2})$ > are commutative diagrams of set-functions > > ```tikz > \usepackage{tikz-cd} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKMsWp0mrdgHEA+uR58QGbHgJFypMTQYs2iEHIBMPcTCgBzeEVAAzAE4QAtkjIgcEJEZ2T9IAFayJjSM9ABGMIwACgJqwiD2WBYAFjiKdo4uiG4eSJoSeuyBCrwZznk0uYjeBVIGADr1DPZoyVim3EA > \begin{tikzcd} > G_1 \arrow[r, "\varphi"] & G_2 \\ > A \arrow[ru, "j_2"'] \arrow[u, "j_1"] & > \end{tikzcd} > \end{document} > ``` > in which $\varphi$ is required to be a [[group homomorphism]]. This is almost a [[coslice category]]; the difference is that we are mixing objects and morphisms of one [[category]] ($\mathsf{Grp}$) with another ($\mathsf{Set}$). $F(A)$, then, will be defined [[terminal objects are unique up to a unique isomorphism|up to isomorphism]] to be (the [[group]] component of) an [[terminal object|initial object]] in $\mathscr{F}^{A}$. That is to say, there should be exactly one morphism from $(j, F(A))$ to another object $(f, G)$, i.e. a unique commutative diagram > > ```tikz > \usepackage{tikz-cd} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADEAKAQQEoQAvqXSZc+QigCM5KrUYs2AcUHCQGbHgJEyk2fWatEIboNkwoAc3hFQAMwBOEALZIATNRwQk0uQba2QagY6ACMYBgAFUU0JEHssCwALHBU7RxdEMhBPb2p9BSMAHUL6ezRErFSQB2c3Dy9MvPlDEAArUwEgA > \begin{tikzcd} > F(A) \arrow[r, "\varphi"] & G \\ > A \arrow[ru, "f"'] \arrow[u, "j"] & > \end{tikzcd} > \end{document} > ``` > > which matches the [[universal property]] stated in the definition above. > > **2.** That $\text{im }R$ is a [[group]] is rather evident. [[associative|Associativity]] follows e.g. from Aluffi exercise 5.4. The empty word $e=()$ is identity, since $we=w=ew$ (no reduction necessary). Finally, given a reduced word $w$, we can construct an inverse by reversing the order of letters of $w$ and replacing each $a\in A$ by $a ^{-1} \in A'$ and each $a ^{-1}$ by $a$. > > **3.** We must show $(j, \text{im }R)$ satisfies the [[universal property]] (at which point we'll write $F(A)=\text{im }R$). This is also rather evident. Any function $f:A \to G$ extends uniquely to a function $\varphi: \text{im }R\to G$, determined by the homomorphism condition and that the diagram commutes, which fixes its value on one-letter words $a \in A$ (as well as on $a^{-1} \in A'$). > > If $f:A \to G$ is any function, we can extend $f$ to a (set-)function $\tilde{\varphi}: W(A) \to G$ > by insisting that on one-letter words ($a$ or $a^{-1}$ for $a \in A$), $\tilde{\varphi}(a)=f(a), \tilde{\varphi}(a ^{-1})=(f(a))^{-1},$ > and that $\tilde{\varphi}$ respects juxtaposition: $\tilde{\varphi}(ww')=\tilde{\varphi}(w) \tilde{\varphi}(w') \text{ for any two words }w,w' \text{ in } W(A).$ > Notice that [[reduced word|reduction]] is invisible for $\tilde{\varphi}$, that is, for any [[word on a set|word]] $w$ we have $\tilde{\varphi}\big( R(w) \big)=\tilde{\varphi}(w)$. Therefore, since $\varphi: \text{im }R \to G$ agrees with $\tilde{\varphi}$ on reduced words, we have for $w,w' \in F(A)$ $\varphi(w \cdot w')=\tilde{\varphi}(w\cdot w')=\tilde{\varphi}(R(ww'))=\tilde{\varphi}(ww')=\tilde{\varphi}(w)\tilde{\varphi}(w')=\varphi(w)\varphi(w').$ > So $\varphi$ is a [[group homomorphism|homomorphism]]. > > ^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 > ```