Free groups are torsion-free

Given the free group \(FX\) on a set \(X\), it is the case that \(FX\) has no tor­sion el­e­ments. That is, ev­ery el­e­ment has in­finite or­der ex­cept for the iden­tity el­e­ment.

Proof

Re­call that we can view ev­ery el­e­ment of the free group as be­ing a freely re­duced word whose let­ters are el­e­ments of \(X\). Also the group op­er­a­tion is “con­cate­na­tion of words, fol­lowed by free re­duc­tion”. It is cer­tainly in­tu­itively plau­si­ble that if we re­peat­edly con­cate­nate a word with more copies of it­self, and then perform free re­duc­tion, we will never reach the empty word; this is what we shall prove. The can­cel­la­tions in the pro­cess of free re­duc­tion are ev­ery­thing here, be­cause the only way the pow­ers of a word can get shorter (and hence the only way the pow­ers of a word can ever come to the empty word) is by those can­cel­la­tions. So we are go­ing to have to analyse the be­havi­our of words at their start and ends, be­cause when we take pow­ers of a word, the start and end are the places where can­cel­la­tion may hap­pen.

We will say a word is cycli­cally re­duced if it is not only freely re­duced, but also it is “freely re­duced if we ro­tate the word round one place”. For­mally, a freely re­duced word \(a_1 a_2 \dots a_n\) is cycli­cally re­duced if and only if \(a_1 \not = a_n^{-1}\). This cap­tures the idea that “the word doesn’t ad­mit any can­cel­la­tion when we take pow­ers of it”.

some ex­am­ples of words which are cycli­cally re­duced and some which are not

Now, ev­ery freely re­duced word may \(w\) be writ­ten as \(r w^\prime r^{-1}\) where \(r\) is a freely re­duced word and \(w^\prime\) is a cycli­cally re­duced word, and there is no can­cel­la­tion be­tween \(r\), \(r^{-1}\) and \(w^\prime\). This is eas­ily proved by math­e­mat­i­cal in­duc­tion on the length of \(w\): it is im­me­di­ate if \(w\) is already cycli­cally re­duced (let­ting \(r = \varepsilon\) the empty word, and \(w^\prime = w\)), while if \(w\) is not cycli­cally re­duced then it is \(a v a^{-1}\) for some let­ter \(a \in X\) and some freely re­duced word \(v\). But then by the in­duc­tive hy­poth­e­sis (since \(v\) is shorter than \(w\)), \(v\) may be writ­ten as \(r v^\prime r^{-1}\) where \(v^\prime\) is cycli­cally re­duced; so \(w = a r v^\prime r^{-1} a^{-1} = (ar) v^\prime (ar)^{-1}\) as re­quired.

More­over, this de­com­po­si­tion is unique, since if \(r w^\prime r^{-1} = s v^\prime s^{-1}\) then \(s^{-1} r w^\prime r^{-1} s = v^\prime\); but \(v^\prime\) is cycli­cally re­duced so there are only two ways to pre­vent can­cel­la­tion:

  • \(s\) is the iden­tity, where­upon \(v^\prime = r w^\prime r^{-1}\) is cycli­cally re­duced, so (by freely-re­duced­ness of \(w = r w^\prime r^{-1}\)) we have \(r = e\) as well, and \(v^\prime = w^\prime = w\).

  • \(s\) is not the iden­tity but it en­tirely can­cels with some­thing in \(r^{-1}\); hence \(s\) is a sub-word of \(r\). But by sym­me­try, \(r\) is a sub-word of \(s\), so they are the same (be­cause with­out loss of gen­er­al­ity \(r\) is also not the iden­tity); and there­fore \(v^\prime = w^\prime\).

Fi­nally, to com­plete the proof that the free group is tor­sion-free, sim­ply take a pu­ta­tive word \(w\) whose or­der \(n\) is finite. Ex­press it uniquely as \(r w^\prime r^{-1}\) as above; then \((rw^\prime r^{-1})^n = r (w^\prime)^n r^{-1}\) which is ex­pected to be­come the empty word af­ter free re­duc­tion. But we already know there is no can­cel­la­tion be­tween \(r\) and \(w^\prime\) and \(r^{-1}\), so there can­not be any can­cel­la­tion be­tween \(r, (w^\prime)^n, r^{-1}\) ei­ther, and by the cycli­cally-re­duced na­ture of \(w^\prime\), our power \(r (w^\prime)^n r^{-1}\) is ac­tu­ally freely re­duced already. Hence our power of the word is em­phat­i­cally not the empty word.

Parents:

  • Free group

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