Show that every non-identity element in a free group \(F\) has infinite order.
Solution
Let \(F=F(X)\) be free on \(X\), and let \(1\ne g\in F\). We show that \(g^m\ne1\) for every integer \(m\ne0\).
1. There exist reduced words \(u\) and \(v\) such that\[ g=uvu^{-1}, \]where \(v\) is nonempty and cyclically reduced.
Start with the reduced word for \(g\). If its first letter is inverse to its last letter, remove those two letters and record the first letter in \(u\). Repeating this operation must terminate because the word is finite. The remaining word \(v\) is nonempty, since otherwise the original reduced word would represent the identity, and its first and last letters are not inverse. Thus \(v\) is cyclically reduced and \(g=uvu^{-1}\).
2. For every integer \(m\ge1\), the word \(v^m\) is reduced and nonempty.
Each copy of \(v\) is reduced. At a junction between two consecutive copies, the last letter of one copy is followed by the first letter of the next. Since \(v\) is cyclically reduced, these letters are not inverse, so no cancellation occurs at any junction. Hence \(v^m\) is a reduced word of positive length.
3. For every integer \(m\ge1\), one has \(g^m\ne1\).
Using 1, \[ g^m=(uvu^{-1})^m=uv^m u^{-1}. \] If \(g^m=1\), conjugating by \(u^{-1}\) and \(u\) would give \(v^m=1\), contradicting 2 because a nonempty reduced word in a free group does not represent the identity.
4. Therefore \(g\) has infinite order.
3.Negative exponents add no further possibility since \(g^m=1\) with \(m<0\) implies\(g^{-m}=1\).