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In Algebra: Chapter 0, one learns two definitions of free groups associating with sets.

  1. Let $A$ be a set, the free group of $A$, $F(A)$ is the initial object in the category $\mathcal{C}$, where \begin{equation} \operatorname{Obj}(\mathcal{C})=\{A\xrightarrow{g}G\}, \end{equation}where the codomain $G$ are groups, and \begin{equation} \operatorname{Hom}(\mathcal{C})=\{\text{Commutative Diagrams } A\xrightarrow{g_1}G_1\xrightarrow{\phi}G_2\xleftarrow{g_2}A\} \end{equation} where $\phi$ are group homomorphisms.

  2. Also $F(A)$ has the concrete construction with elements being non-redundant words with alphabet $A$, and the group multiplication being juxtaposition and reduction.

But what good are free groups? Why are they useful? Wikipedia says they are useful in topology, but does not explain why explicity.

Can someone give some examples that an undergraduate can understand? I am asking about examples that can show the usefulness of this abstract construction. So I guess the identification of $\mathbb{Z}=F({a})$ does not really count.

Mike Pierce
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Hui Yu
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    What good is a free dinner? Really, do you have to ask? – GeoffDS Nov 17 '12 at 16:00
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    Well, I'm personally a big fan of the integers--a.k.a. the free group with one generator. – Cameron Buie Nov 17 '12 at 16:01
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    Have you ever stopped to think what it means to give a presentation of a group by generators and relations? You'll find that you will need the concept of free group in order to make sense of this rigorously. – Zhen Lin Nov 17 '12 at 16:02
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    @ZhenLin That is a good point! But this is actually only a technicality. Is there some good examples that we use free groups to solve problems? – Hui Yu Nov 17 '12 at 16:09
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    @HuiYu, Just knowing what a free group wont solve any complicated problems. It's just a natural concept that arises in group theory. – sperners lemma Nov 17 '12 at 16:12
  • This doesn't answer your question but it may be of interest: http://www.springerlink.com/content/ll61220g664q4475/ – Derek Allums Nov 17 '12 at 16:26
  • Also, how much topology do you know? In particular, do you know how to find the fundamental group of the wedge of circles? – Derek Allums Nov 17 '12 at 16:27

3 Answers3

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As an example of whar @Berci noted; we know that:

Every group $G$ is a quotient of a free group.

Theorem above let us to identified $G$ with $\frac{F}{R}$ where in $F$ is a free group and $R$ is a normal subgroup of it. Now if we denote $X$ as a base for $F$ and $\delta$ as a family of words (like a sequence of alphabets above), then $X$ is known as the generators of $G$ and $\delta$ as its relations. Here, we propose a new symbol as $G=\langle X|R \rangle$ and we call it a presentation of $G$. As I have worked some problems in this kind of structures, without knowing a presentation of a group some beautiful facts about that group may be neglected. How can we introduce $A_4$ easier. How can we work with $S_3$ when a student forget its certain permutations? The Free groups and their great theorems help us to work with such these groups easier. Indeed $A_4=\langle a,b|a^2=b^3=(ab)^3=1\rangle$ and $S_3=\langle a,b|a^2=b^3=(ab)^2=1\rangle$. It is easy to memorize, at least for me.

Mikasa
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Usually, for algebraic structures,

Any structure is a quotient of a free such structure.

If we are given abstractly operation symbols $\mu,\nu,+,\cdot,...$, and a positive number as the arity of the operation (e.g. '$+$' is a binary symbol), then one can define formal terms, built up by pregiven variables (associated to the elements of the given set $A$) and these operation symbols. Basically these terms are just 'strings' like $\mu(a+b,c\cdot d, e)$ if $\mu$ is ternary operation symbol.

The main thing is, that the set of terms naturally admits a structure of the given type, the operation symbols act just formally. And this is the free algebra of the given type.

If, in addition, some rules are present (axioms for the given structures, like associativity for groups) , then these rules should also be applied to the set of the terms, considering a quotient of it, e.g. where the strings $(a\cdot b)\cdot c$ and $a\cdot (b\cdot c)$ are identified with each other. (Hence, if associativity of the operation is imposed, it is enough to consider the finite sequences like $(a,b,c,a,a,c)$.)

So, free groups, in this sense, are so general, that every group is 'contained' (as a quotient) of them. On the other hand, every group can be embedded to a symmetric group. So, symmetric groups have similar important role (at least, in this view).

In topology, for example, the fundamental group of $n$ circles glued together at one point is the free group on $n$ generators.

Free groups also may provide easy counterexamples of statements like 'identities $P,Q$ on groups imply identity $R$'. If this is not true, then the quotient 'by $P$ and $Q$' of a suitable free group must be a counterexample.

Berci
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Here's a neat application which I originally learned from a paper of Bergman: given a group $G$ and letting $F_k$ be the free group on $k$ elements, we can think of $G*F_k$ as being $G$ augmented with $k$ "generic" elements (and this can be generalized to other categories of algebraic structutres with the free product being replaced by coproduct). Given some additional structure on $G$ which "lifts along" homomorphisms out of $G$, we can use this to get nontrivial information. For example, the Bergman paper linked above shows that for an automorphism $\alpha$ of a group $G$, the following are equivalent:

  • There is an assignment $Hom(G,-)\rightarrow Aut(-):f\mapsto \beta_f$ such that $\beta_{f_2}h=h\beta_{f_1}$ for any $h: H_1\rightarrow H_2$, $f_1: G\rightarrow H_1$, $f_2: G\rightarrow H_2$ with $hf_1=f_2$.

  • $\alpha$ is an inner automorphism of $G$, that is, for some $g\in G$ we have $\alpha:x\mapsto gxg^{-1}$.

It actually says more than this, the issue being that multiple $\beta_-$s may correspond (= evaluate at $\mathit{id}$) to the same $\alpha$. It turns out that the $\beta_-$s with the above properties are in really nice correspondence with the elements of $G$ themselves: a given $\beta_-$ must be of the form $f\mapsto (x\mapsto f(g)xf(g)^{-1})$ for some $g\in G$.

Noah Schweber
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