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Negative immersions for one-relator groups

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 Added by Henry Wilton
 Publication date 2018
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and research's language is English




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We prove a freeness theorem for low-rank subgroups of one-relator groups. Let $F$ be a free group, and let $win F$ be a non-primitive element. The primitivity rank of $w$, $pi(w)$, is the smallest rank of a subgroup of $F$ containing $w$ as an imprimitive element. Then any subgroup of the one-relator group $G=F/langlelangle wranglerangle$ generated by fewer than $pi(w)$ elements is free. In particular, if $pi(w)>2$ then $G$ doesnt contain any Baumslag--Solitar groups. The hypothesis that $pi(w)>2$ implies that the presentation complex $X$ of the one-relator group $G$ has negative immersions: if a compact, connected complex $Y$ immerses into $X$ and $chi(Y)geq 0$ then $Y$ is Nielsen equivalent to a graph. The freeness theorem is a consequence of a dependence theorem for free groups, which implies several classical facts about free and one-relator groups, including Magnus Freiheitssatz and theorems of Lyndon, Baumslag, Stallings and Duncan--Howie. The dependence theorem strengthens Wises $w$-cycles conjecture, proved independently by the authors and Helfer--Wise, which implies that the one-relator complex $X$ has non-positive immersions when $pi(w)>1$.

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Previously, the authors proved that the presentation complex of a one-relator group $G$ satisfies a geometric condition called negative immersions if every two-generator, one-relator subgroup of $G$ is free. Here, we prove that one-relator groups with negative immersions are coherent, answering a question of Baumslag in this case. Other strong constraints on the finitely generated subgroups also follow such as, for example, the co-Hopf property. The main new theorem strengthens negative immersions to uniform negative immersions, using a rationality theorem proved with linear-programming techniques.
We show that any one-relator group $G=F/langlelangle wranglerangle$ with torsion is coherent -- i.e., that every finitely generated subgroup of $G$ is finitely presented -- answering a 1974 question of Baumslag in this case.
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