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Are there any procedures which given a nonnegative nondecreasing function on the integers will construct a finitely generated group with the same growth up to the usual equivalence of growth functions? Can this at least be done for nice functions? For example Bergman gives an explicit construction of semigroups of any growth between quadratic and cubic.

Also, is there an algorithmic way to do this if your function is recursive and given as input by a Turing machine?

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    $\begingroup$ Why is this question so terrible? Constructing groups with intermediate growth of a somewhat prescribed type is very difficult. We don't even know if there is a smallest intermediate growth. $\endgroup$ Dec 8, 2013 at 2:23
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    $\begingroup$ You might like this classical paper of Bergman math.berkeley.edu/~gbergman/papers/unpub/growth.pdf $\endgroup$ Dec 8, 2013 at 2:40
  • $\begingroup$ @BenjaminSteinberg Do you feel that the reason for closure is invalid? $\endgroup$
    – Todd Trimble
    Dec 8, 2013 at 3:02
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    $\begingroup$ @BenjaminSteinberg : In case you miss my meta comment, here it is again. I agree that one could ask a reasonable question about prescribing the growth of groups. However, this is not it. For instance, I don't know what the OP means by "manifold", and I don't know exactly what the OP means with regards to a growth function, e.g. what's the equivalence relation? For instance, as written he might have a specific function and want a group with a specific finite generating set realizing that specific function on the nose. Someone (you or the OP) should rewrite the question before it is reopened. $\endgroup$ Dec 8, 2013 at 3:16
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    $\begingroup$ Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomial growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growth are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is subexponential growth, and the first examples with growth determined up to equivalence were obtained quite recently by Bartholdi and Erschler (Inventiones arxiv.org/abs/1011.5266, 2012, see also their more recent arxiv.org/abs/1110.3650) $\endgroup$
    – YCor
    Dec 8, 2013 at 17:12

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Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomially bounded growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growths are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is intermediate growth, and the first examples with growth determined up to equivalence were obtained quite recently by Bartholdi and Erschler (Inventiones, arxiv 1011.5266, 2012, see also their more recent arxiv 1110.3650). Although the growth of the first Grigorchuk group $\Gamma$ is not yet known up to equivalence, $\Gamma$ is used (in a permutational wreath product construction) in the construction of the Bartholdi-Erschler groups.

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    $\begingroup$ In particular it is not known whether there is a group with growth function equivalent to $e^{\sqrt{n}}$. $\endgroup$ Dec 8, 2013 at 22:22

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