Formal definition of the free group

Finally, the map is surjective: we can make any member of the free group by “converting the appropriate reduced word into a function”. Indeed, the free group is generated by the \(\rho_x\) and \(\rho_{x^{-1}}\) for \(x \in X\), so every element is some \(\rho_{x_1} \dots \rho_{x_n}\) for some selection of \(x_1, \dots, x_n \in X \cup X^{-1}\). Note that \(x_1 \dots x_n\) need not necessarily be freely reduced as a word at the moment; but if it is indeed not freely reduced, so some \(x_i, x_{i+1}\) cancel each other out, then removing that pair completely doesn’t change the function \(\rho_{x_1} \dots \rho_{x_n}\). For example, \(\rho_{x_1} \rho_{x_1^{-1}} \rho_{x_2} = \rho_{x_2}\). Hence the process of “performing one step of a free reduction” (i.e. removing a cancelling pair) doesn’t change the member of the free group as a function; and since each such removal makes the word shorter, it must eventually terminate. It remains to show that it doesn’t matter in what order we remove the cancelling pairs; but that is immediate because we’ve already shown that our “conversion” process is injective: we started with a member of the free group, so if it corresponds to a freely reduced word then it corresponds to a unique freely reduced word. Since we’ve just shown that it does indeed correspond to a freely reduced word (by repeatedly removing cancelling pairs), we are done.

The above shows that the free group can be considered just to be the set of reduced words.