TITLE: Different ways of constructing the free group over a set.
QUESTION [17 upvotes]: This could be too broad if we're not careful. I'm sorry if it ends up that way.

Let's put together a list of different constructions of the free group $F_X$ over a given set $X$.

It seems to be one of those things a lot of people know about (and use implicitly) but whose constructions can be so tedious, it's hard to get hold of at first. The main problem, as Lee Mosher reminds us, seems to be associativity.
The Wikipedia page (linked to above) goes some way into listing some useful perspectives. I'm impressed by this intuitive summary from Wolfram:

A group is called a free group if no relation exists between its group generators other than the relationship between an element and its inverse required as one of the defining properties of a group.

But this is not exactly a construction nor is it strictly the free group over a set.
Here's a brief list of what I have so far:

An alternative approach to constructing the free group.

This starts with the "standard" construction  using finite strings over $\mathcal{X}=X\cup X'$ then asks for an opinion on quotienting by some equivalence relation. 

Can a free group over a set be constructed this way (without equivalence classes of words)?

The free group is constructed as the left adjoint of the composition of certain forgetful functors. This one is interesting in that it goes via the category InvMon of monoids with involutions as objects and involution-preserving homomorphims as morphisms.  
Martin Brandenburg also gives quite a concise one in the comments there, so I invite him to elaborate on that here :)

Magnus et al. in "Combinatorial Group Theory: Presentation of Groups in Terms of Generatorators and Relations" manage to define group presentations first of all in their opening chapter.
W. Ledermann in "Introduction to Group Theory" in $\S V$ gives a very standard construction by first defining finite words over $X$, then what it means to be a reduced word, then taking $F_X$ as the reduced words over $X$ under (reduced) concatenation.


The above list is not at all exhaustive (of what I know) and there's bound to be some overlap. Personally I would be interested to see explicit use of Universal Algebra, Semigroup Theory, and Category Theory. The reason for the latter should be clear from the above; as for the first two, see

Page 68 onwards of "A Course in
Universal Algebra" by Burris et al. and
Chapter 2 of "Nine Chapters on the Semigroup Art" by A.J. Cain.


Feel free to give more details on those already listed here.

REPLY [3 votes]: Although this is an old question, this construction recently came up by reference in another question by Shaun, and it does not appear to be included above, so I figure I can add it (having checked with OP first).
Here's a categorical construction using products; it is actually related to the adjoint functor construction, in that it follows the ideas of Freyd's General Adjoint Functor Theorem. It can be carried out with any class of algebras that is closed under products and has a notion of "subobject generated". So it works in most "nice" categories of algebras (in the sense of Universal Algebra): abelian groups, $R$-modules, commutative rings, associative rings, etc.
Let $X$ be a set. A free group on $X$ is a pair $(F,u)$, where $F$ is a group, $u\colon X\to F$ is a set map, and the pair has a universal property: for every group $G$ and every set map $v\colon X\to G$, there exists a unique group homomorphism $\phi\colon F\to G$ such that $v = \phi\circ u$.
The idea is this: consider all pairs $(G_v,v)$ where $G_v$ is a group, and $v\colon X\to G_v$ is a set map. Take the product
$$P=\prod_{(G_v,v)} G_v,$$
of all such $G$, indexed by the pairs. The maps $v_G\colon X\to G$ induce a set map $u\colon X\to P$. Let $F$ be the subgroup of $P$ generated by $u(X)$. Then $F$ and $u$ will have the desired property, because given any group $G$ and set map $v\colon X\to G$, we obtain the relevant morphism $F\to G$ by simply taking the restriction of the projection $\pi_{(G_v,v)}\colon P\to G$ to the subgroup $F$. And because $F$ is generated by $u(X)$, this will provide the uniqueness. Voila! As if by magic, we obtain a group with the correct universal property.
However, there is a technical issue with the idea above: we can't construct that product, because the collection of all pairs $(G_v,v)$ is a proper class, not a set, and we cannot take a cartesian product indexed over a proper class. So we need to pare down this collection to something manageable (i.e., to a set).
The first observation we can make is that if $f\colon G\to H$ is an isomorphism, and $v\colon X\to G$ is a set map, then we do not need both $(G,v)$ and $(H,f\circ v)$ in our collection: we can just take $(G,v)$, and if we are presented with $H$ and $f\circ v$, we project to $G$ and then use $f$ to get all the way to $H$. So we really only need one group from each isomorphism class, plus all set maps from $X$ to that group (and that collection, for a fixed group $G$, is a set).
However, this is still not small enough, since there are groups of arbitrarily large cardinality, and thus even taking just one group from each isomorphism class we would still have a proper class. So we will not go down this road in the end (turns out to not matter).
The second observation to make is that in fact, given $(G,v)$, we don't really need all of $G$: we just need $\langle v(X)\rangle$, the subgroup generated by $v(X)$. Because the image of $F$ in $G$ will be generated by $v(X)$, so the image will lie inside that subgroup.
This is a good observation, because that bounds the size of generating sets we need to consider. And now we have an easy upper bound for how big the groups we need to consider are:
Lemma. Let $G$ be a group, and $S\subseteq G$ a subset of $G$. If $G=\langle S\rangle$, then $|G|\leq |S|\aleph_0$.
Proof. We know that $\langle S\rangle$ consists of finite products of elements of $S$ and their inverses. The set of such products has at most $|S|\aleph_0$ elements, so $G$ has at most that many elements. $\Box$
So, here's how we get around the technical issue: fix a set $M$ of cardinality $|X|\aleph_0$. We will consider all pairs $(G_v,v)$ such that

$G_v$ is a group;
$v\colon X\to G_v$ is a set map;
$G_v$ is generated by $v(X)$;
The underlying set of $G_v$ is contained in $M$;

Call the collection of all such pairs $\mathcal{S}$. It is now easy to verify that this is a set.
This is what is sometimes called a "solution set". It has the following property:

Given any group $K$ and set map $w\colon X\to K$ such that $K=\langle w(X)\rangle$, there exists a pair $(G_v,v)\in\mathcal{S}$ and an isomorphism $\phi\colon G_v\to K$ such that $w = \phi\circ v$.

So now we proceed as outlined above: Let $P$ be the product over all elements of $\mathcal{S}$,
$$P = \prod_{(G_v,v)\in \mathcal{S}} G_v.$$
Let $\pi_v\colon P\to G_v$ be the projection onto the $(G_v,v)$th coordinate. Because $\mathcal{S}$ is a set, this product exists and makes sense.
The underlying set of $P$ is the cartesian product of the $G_v$. Thus, the functions $v\colon X\to G_v$ induce a unique set map $u\colon X\to P$ such that $\pi_v\circ u = v$ for all $(G_v,v)\in\mathcal{S}$.
Now, we "remember" that $P$ is in fact a group, so let $F=\langle u(X)\rangle$ be the subgroup of $P$ generated by image of $u$. I claim that $(F,u)$ is the free group on $X$.
Indeed, let $H$ be any group, and let $w\colon X\to H$ be a set map. Then $K=\langle w(X)\rangle$ is a group generated by $w(X)$, and hence by the solution set property of $\mathscr{S}$, there exists a $(G_v,v)\in\mathcal{S}$ and an isomorphism $\phi\colon G_v\to K$ such that $w=\phi\circ v$. Let $\iota\colon K\to H$ be the inclusion map.
Then $\iota\circ\phi\circ\pi_v\colon F\to G_v\to K\hookrightarrow H$ is a morphism from $F$ to $H$, and
$$\begin{align*}
(\iota\circ\phi\circ\pi_v)\circ u &= \iota\circ\phi\circ (\pi_v\circ u)\\
&= \iota\circ\phi\circ v\\
&= w
\end{align*}$$
(because $\phi\circ v$ is equal to the co-restriction of $w\colon X\to H$ to $K$).
Thus, there exists a morphism $f\colon F\to H$ such that $f\circ u = w$. To prove uniqueness, we note that if $g\colon F\to H$ also satisfies $g\circ u=w$, then for every $x\in X$ we have $g(u(x)) = w(x) = f(u(x))$, so $g$ and $f$ agree on $u(X)$. Therefore, they agree on $\langle u(X)\rangle$; by construction, this equals $F$, so $f=g$.
Hence $(F,u)$ has the universal property of the free group on $X$, and hence is the free group on $X$ (up to unique isomorphism). $\Box$