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.


  • Free group

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