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?