# Every torsion-free group is an abelian group?

Is it true that every torsion-free group (i.e. a group where the only element of finite order is the identity) is abelian?

Free groups are torsion free but not abelian!

• As a nice aside, one way to see this is to consider the Cayley graph of a finitely generated free group. Such a Cayley graph is a tree so has no cycles. Hence repeating any path (I.e raising an element to any power) will never form a loop in the graph. Hope this helps. – Zestylemonzi Jul 27 '16 at 14:39
• There are two abelian free groups. Free groups on two or more (free) generators aren't abelian. – Daniel Fischer Jul 27 '16 at 19:07
• Yup, cheers for pointing that out. – Zestylemonzi Jul 27 '16 at 19:53

The Heisenberg group is torsion-free but not abelian. It is the set of matrices of the form \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}

It is torsion-free because $$\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}^n = \begin{pmatrix} 1 & na & {n \choose 2}ab+nc\\ 0 & 1 & nb\\ 0 & 0 & 1\\ \end{pmatrix}$$

It is not abelian because $(AB)_{13} \ne (BA)_{13}$.