TITLE: What good are free groups?
QUESTION [14 upvotes]: In Algebra: Chapter 0, one learns two definitions of free groups associating with sets.


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.
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.

REPLY [14 votes]: 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.