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

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Free groups are torsion free but not abelian!

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    $\begingroup$ 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. $\endgroup$ – Zestylemonzi Jul 27 '16 at 14:39
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    $\begingroup$ There are two abelian free groups. Free groups on two or more (free) generators aren't abelian. $\endgroup$ – Daniel Fischer Jul 27 '16 at 19:07
  • $\begingroup$ Yup, cheers for pointing that out. $\endgroup$ – Zestylemonzi Jul 27 '16 at 19:53
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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 $(AB)_{13} \ne (BA)_{13}$.

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