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

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?

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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.) – HJRW Oct 10 '12 at 11:05
@HW: Yes, of course: $5/11 > 1/4$. – Mark Sapir Oct 10 '12 at 12:39
Mark - I agree with your inequality! Hence my wanting to check I'd remembered the definition correctly. – HJRW Oct 10 '12 at 12:51
The question should probably be about 2/11 which is between 1/5 and 1/6. – Denis Osin Oct 10 '12 at 15:24
Many thanks for your helpful comments! Denis, would it have exponential growth if it was 2/11? – Elisabeth Fink Oct 11 '12 at 10:29

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".
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 '12 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 '12 at 12:17