# 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”.