Definition 0. Let $R$ denote a commutative semiring with $0$ and $1$. By an $R$-monoid, I mean a monoid $M$ equipped with an action $R \times M \rightarrow M$ denoted $r,m \mapsto m^r$, satisfying the following axioms.

  1. $m^1 = m$
  2. $(m^r)^s = m^{rs}$
  3. $m^0 = 1$
  4. $m^{r+s} = m^r m^s$

Remark. If $R$ is a ring, then every $R$-monoid is a group, because $1 = m^{0} = m^{-1+1} = m^{-1} m.$


  1. An $\mathbb{N}$-monoid is just a monoid.
  2. A $\mathbb{Z}$-monoid is just a group.
  3. The set $\mathbb{R}_{>0}$ can be regarded as an $\mathbb{R}$-monoid (or a $\mathbb{Q}$-monoid, if we wish), where the law of composition is $(p,q) \mapsto pq$ and the action is $x,p \mapsto p^x.$


  1. What are some examples of non-commutative $\mathbb{Q}$-monoids and/or $\mathbb{R}$-monoids?
  2. What is the usual terminology for $R$-monoids?
  3. Is there any literature surrounding them? References appreciated.

2 Answers 2


A nice example. Goes back to Cayley, 1860 [1].

Formal power series of the form $$ f(x) = x + a_1x^2+a_2x^3+a_3x^4+\cdots $$ with real coefficients. Under composition. Cayley showed how to do "fractional" composites, of real (or even complex) order. The series need not converge, even if $f$ does. That's why I said "formal" power series.

plug: I did a generalization for transseries [2].

[1] A. Cayey, On some numerical expansions. Quarterly Journal of Pure and Applied Mathematics 3 (1860) 366--369

[2] G. Edgar, Fractional iteration of series and transseries. Transactions of the American Mathematical Society 365 (2013) 5805--5832

  • 1
    $\begingroup$ If you restrict to rational coefficients, you get a $\mathbb Q$-monoid. If you work formally subject to $x^n=0$, this example still works, but now it is a finite-dimensional nilpotent Lie group, quite like André's example. $\endgroup$ Jun 8, 2015 at 17:26

I would guess that an $\mathbb R$-monoid is the same thing (once you put extra smoothness axioms on the $\mathbb R$-action) as a Lie group whose exponential map $\mathfrak g\to G$ is a diffeomorphism. The Heisenberg group $$ \left(\begin{matrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \\ \end{matrix}\right) $$ is the simplest non-commutative example of such a group. Take the same example but restrict $*\in \mathbb Q$ to get an example of a $\mathbb Q$-monoid which is not an $\mathbb R$-monoid.

  • 2
    $\begingroup$ For the $\mathbb Q$ version to work it is important that the group to be nilpotent, so that the exponential preserve rational matrices. . . There is a smaller, though perhaps not simpler, non-commutative Lie group, the affine group $ax+b$. The exponential is still a bijection, so it gives an $\mathbb R$-group, but the exponential does not preserve rational elements. It contains $\mathbb Q$-monoids, but they are more complicated. $\endgroup$ Jun 8, 2015 at 17:32

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