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Suppose $G$ has the presentation $\langle t, x_1, x_2, ... | R \rangle$ where each relator in $R$ has the form $t^{-1}x_it = x_j$ for some $i,j$. Does $G$ have an element of order 2?

This is an HNN extension of a free group, if that changes anything.

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By the torsion theorem for HNN-extensions, every element of finite order is conjugate to an element of the base, which in your case is a free group, so the answer is no.

See Lyndon and Schupp's Combinatorial Group Theory.

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  • $\begingroup$ This is unfortunate for at least one author who I'll leave anonymous, because it provides a counterexample to a theorem in one of his books on one relator groups. $\endgroup$
    – JeremyKun
    Commented Apr 16, 2011 at 3:31
  • $\begingroup$ Note that my answer only works when $G$ is in fact an HNN-extension of a free group, which isn't obviously the case for all groups with the kind of presentation you've given. (For instance, the relations could be $t^{-1}x_i t = x_1$ for all $i$, and so $t$ isn't giving you an isomorphism between subgroups.) Could that be the source of the problem? $\endgroup$
    – Autumn Kent
    Commented Apr 16, 2011 at 4:19
  • $\begingroup$ No. It is definitely an HNN extension of a free group. I just posed the more general question because I didn't know of this theorem on HNN extensions, and thought to answer it from a different angle. $\endgroup$
    – JeremyKun
    Commented Apr 16, 2011 at 16:26

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