# 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.

Parents:

• Free group

The free group is “the purest way to make a group con­tain­ing a given set”.

• Group

The alge­braic struc­ture that cap­tures sym­me­try, re­la­tion­ships be­tween trans­for­ma­tions, and part of what mul­ti­pli­ca­tion and ad­di­tion have in com­mon.