# Understanding torsion from a presentation

Let $F_2 = \langle a,b \rangle$ be the free group on two generators, and for each word $w \in F_2$, let $G(w) = \langle a, b \ | \ w \rangle$. Is the following statement true?

$G(w)$ is torsion free if and only if for all $k \geq 2$ and for all $v \in F_2$, $w \neq v^k$

In other words, is it true that $G(w)$ is torsion free unless there is an obvious reason why it is not?

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## 1 Answer

Yes, this is true. It is proved as Theorem 4.12 in "Combinatorial Group Theory", by W. Magnus, A. Karrass and D. Solitar. (It's on p. 266 in the Dover reprint.) Alternatively, see Propositions 5.17 and 5.18 in "Combinatorial Group Theory", by R.C. Lyndon and P.E. Schupp. It is true without the restriction on the rank of the free group, though you can reduce to that case by an embedding. The most straight-forward proof uses induction on the length of the relator.

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I believe the second author's name is "Abraham Karrass", with a K. –  Arturo Magidin Jun 20 '12 at 19:54
Thank you very much James. I felt like this should be known, but I was having trouble finding it! –  Dan Glasscock Jun 21 '12 at 0:55
Whoever looks for those propositions in Lyndon/Schupp could get anguished a little, as this book numbers from scratch theorems and propositions in every chapter. It must be Chapter II, prop's 5.17 - 5.18 (pag. 107 of the SPRINGER Ed.). In fact, the proof (leeeengthy one) only appears in M-K-S, and L-S mentions only "they" proved it, so if someone needs the proof he must go to M-K-S. –  DonAntonio Jun 21 '12 at 2:11
@ArturoMagidin. Indeed it is Karrass. Thanks for catching that. –  James Jun 21 '12 at 21:12
@DonAntonio. Actually, the proof is in Lyndon & Schupp. They state it again later, in Chapter IV, Theorem 5.2. (You're right about the numbering.) A proof along the lines I remembered (much shorter than in MKS) is included at that point. –  James Jun 21 '12 at 21:17
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