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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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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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    @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.2012-06-21