# 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”.

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”.