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The notion of nilpotency passes nicely from groups to semigroups. Define $q_1(x,y)=xy$ and $$q_{i+1}(x,y,z_1,\cdots,z_i)=q_i(x,y,z_1,\cdots,z_{i-1})z_iq_i(y,x,z_1,\cdots,z_{i-1})$$ inductively for all $x,y,z_1,z_2,\cdots$.

A. I. Malcev proved that nilpotent groups of class $c$ are described as the groups satisfying the law $q_c(x,y,z_1,\ldots,z_{c-1})=q_c(y,x,z_1,\ldots,z_{c-1})$. It is natural to define nilpotent semigroups of class $c$ as those satisfying this law.

Question: Do all f.g. nilpotent semigroups have polynomial growth?

Note that for cancellative semigroups the answer is "yes" since nilpotent cancellative semigroups satisfy the Ore's condition and so are group-embeddable.

(My guess is that in general the answer is "no" and there even perhaps exists a counter-example among matrix semigroups.)

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The answer is "no", see Theorem 6.1 in Shneerson, L. M. Relatively free semigroups of intermediate growth. J. Algebra 235 (2001), no. 2, 484–546. The relatively free semigroup in the variety $xyzyx=yxzxy$ ("nilpotent" of class 2) with at least 3 generators has intermediate growth ($\sim 2^{\sqrt{n}}$).

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  • $\begingroup$ Fantastic -- even intermediate growth!! Thanks a lot, Mark!!! $\endgroup$
    – Victor
    Commented May 31, 2011 at 5:59
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    $\begingroup$ For higher nilpotency class, the relatively free groups must have exponential growth. It should follow from my paper on Burnside properties in semigroups (there is a reference in Shneerson's paper) but it is not written anywhere as far as I know. $\endgroup$
    – user6976
    Commented May 31, 2011 at 7:37
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Somewhat related: Th. (Grigorchuk) : A finitely generated cancellative semigroup has polynomial growth if and only if it has a virtually nilpotent group of left quotients.

Ref.: Grigorchuk, R.I.: Cancellative semigroups of power growth. Math. Notes 43(3) (March 1988) 175–183. (The original version: Р. И. Григорчук, “О полугруппах с сокращениями степенного роста”, Матем. заметки, 43:3 (1988), 305–319 )

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  • $\begingroup$ This is essentially already contained in Victor's post without the reference to Grigorchuk. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 2, 2014 at 23:22

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