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I am reading an introduction to growth of groups. The notions of polynomial and superpolynomial growth are introduced, as are exponential and subexponential growth.

I can prove that the growth of a group is always either exponential or subexponential (it is exercise 1.6). However, there seems to be no mention of an analogous result for (super)polynomial growth (i.e. the growth of a group is always either polynomial or superpolynomial).

There exist strictly increasing functions which grow faster than polynomially but are not superpolynomial (this is pretty clear; a more detailed explanation can be found in the second section of this document), but I do not know whether these occur as the growth function of some group.

The thesis of a Nick Scott claims to prove that every group grows either polynomially or superpolynomially, but I don't see it (it is in subsection 1.4.1, on p.12; it seems to me the proof assumes that the limit $\log(\beta(k))/\log(k)$ exists, but I don't know why).

So my question is: does every group grow either polynomially or superpolynomially?

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2 Answers 2

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Yes. This is a result of Grigorchuk, see his Mittag-Leffler notes.link text (Milnor's problem on the growth of groups and its consequences, available on line for free; see page 28).

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    $\begingroup$ seems to be p. 19, Cor. 8.6 $\endgroup$
    – Ian Agol
    Commented Apr 26, 2012 at 2:59
  • $\begingroup$ @Agol, yes, true, but the same result is mentioned in a broader "Gap" context on pp 28-29. $\endgroup$
    – Igor Rivin
    Commented Apr 26, 2012 at 3:57
  • $\begingroup$ Thank you very much! It is indeed mentioned in cor. 8.6 and on page 28-29. On page 18 there is mention of work by van der Dries and Wilkie also proving it. $\endgroup$ Commented Apr 26, 2012 at 8:43
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    $\begingroup$ The link in the message is broken. Here's the arxiv link: arxiv.org/abs/1111.0512 $\endgroup$
    – YCor
    Commented Jan 20, 2016 at 23:37
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    $\begingroup$ By the way Grigorchuk in this survey does not claim this result as his own result. As far as I remember, I think it's folklore that it follows from Gromov's original proof. $\endgroup$
    – YCor
    Commented Jan 20, 2016 at 23:41
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It is interesting that another asymptotic invariant of a group, the Dehn function, can be arbitrary large (even non-recursive) but still bounded by a polynomial on an infinite set. In fact the polynomial can be made quadratic. See this paper.

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