Let $A$ be a set. We wish to construct the free group $F(A)$. It seems that this (invariably?) starts out like this:
Let $A'$ be a copy of $A$, and let $\mathscr A=A\cup A'$.
Let $\mathscr L$ be the set of finite strings of elements of $\mathscr A$.
Then it seems the usual approach does this:
Show that each element of $\mathscr L$ can be reduced to a unique irreducible form. Show that concatenation followed by reduction is a group operation on the set of irreducible forms, forming a group $F(A)$. Prove that this satisfies the universal property of a free group.
Possible alternative?
Having defined $\mathscr L$, define an equivalence relation $\sim$ by letting $s\sim t$ iff there is some finite sequence of reductions and/or expansions that transform $s$ into $t$. Let $F(A)=\mathscr L/\sim$.
This appears to make it really easy to prove that $F(A)$ is a group, and I believe it is still easy to see it is the free group, but I don't have a clue how to prove that each element has a unique irreducible representative, which would, I believe, make this a nice clean drop-in replacement for what appears to be the standard approach.
Any tips?
A topologist would say: Take the fundamental group of the wedge of circles of the given cardinality (rank of the group you want to get). However, this assumes that you know what fundamental groups are (and van Kampen's theorem).
Edit: If you are looking for purely algebraic approaches, take a look also at the answer
Slicker construction of the free product of groups
since the free product of $c$ copies of infinite cyclic groups satisfies is a free group of rank $c$ (one can see this via the universal property for both). I do not know how this answer compares with DonAntonio's.