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I came across this rather week small cancellation condition $C'\left(\frac{5}{11}\right)$ of a group $G$. It has been proved that $C'\left(\frac16\right)$ is enough for $G$ to contain free subgroups. I was therefore wondering if $\frac{5}{11}$ is maybe enough to still have exponential growth.

Does anyone know of any related papers or results?

flag	asked Oct 10 at 10:59 Elisabeth Fink 167●4

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Doesn't $\mathbb{Z}^2$ satisfy $C'(5/11)$? Or have I misremembered the definition? (You should probably give the definition.) – HW Oct 10 at 11:05

@HW: Yes, of course: $5/11 > 1/4$. – Mark Sapir Oct 10 at 12:39

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Mark - I agree with your inequality! Hence my wanting to check I'd remembered the definition correctly. – HW Oct 10 at 12:51

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The question should probably be about 2/11 which is between 1/5 and 1/6. – Denis Osin Oct 10 at 15:24

Many thanks for your helpful comments! Denis, would it have exponential growth if it was 2/11? – Elisabeth Fink Oct 11 at 10:29

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Mark Sapir · Accepted Answer · 2012-10-10 15:03:10Z UTC

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Every finitely presented group has presentation satisfying $C'(1/5)$. Note that $1/5 < 5/11$. See the book by Olshanskii's book "Geometry of defining relations of groups".

edited Oct 10 at 15:03

answered Oct 10 at 12:41

Mark Sapir
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It is A.I.Golberg's theorem (1978). The proof can be found in Olshanskii's book, but not in the book of Lyndon and Schupp (1977), I think. – Anton Klyachko Oct 10 at 14:27

@Anton: You are right! – Mark Sapir Oct 10 at 15:02

Many thanks for your answer. The groups I am looking at are infinitely presented. Are there any partial results known for such cases? – Elisabeth Fink Oct 11 at 10:51

If the group is not finitely presented, the situation is different. At least you know, by Gromov's theorem, that the group cannot have polynomial growth. I do not know if Grigorchuk group or other groups of intermediate growth can have $C'(5/11)$-presentation. You can check the Lysenok presentation of Grigorchuk group. – Mark Sapir Oct 11 at 11:48

Lysenok's presentation does not satisfy $C'(5/11)$ because of the relations $a^2=1, b^2=1, c^2=1, d^2=1$. One probably needs to look at other groups. There are torsion-free groups of intermediate growth constructed in R. I. Grigorchuk. Degrees of growth of p-groups and torsion-free groups. Mat. Sb. (N.S.), 126(168)(2):194–214, 286, 1985. – Mark Sapir Oct 11 at 12:17

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Does $C'\left(\frac{5}{11}\right)$ imply exponential growth?

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