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Suppose that $M$ is a monoid, and $A$ an $M$-set with action $\alpha: M\times A \to A$.

Suppose that there is a non-empty subset $X\subseteq A$ with the property that:

  • For any $f: X \to A$ there is a unique $i\in M$ such that $f(x) = \alpha(i,x)$ for all $x\in X$

Question: is an $X$ with this property unique if it exists? Must such a set always exist? (Also: is there a standard name for this concept?)

Andrew Bacon
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  • For $M$ a group this seems to almost never exist. $X$ must consists of at most one element from each orbit, so that the inclusion $X\to A$ has the desired property. However, if $f$ sends any element out of its orbit, then no such $i$ will exist. – Nex Sep 16 '17 at 19:24
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    If $M={e}$ is the trivial monoid and $|A|\geq 2$, then no such set $X$ exists. – M. Winter Sep 16 '17 at 19:25
  • Awesome, thanks! Do you know if $X$ is unique if it exists? The application I have in mind is the monoid of substitutions on propositional letters acting on a propositional language: there $X$ exists (it is the set of propositional letters) and is unique. So I think the notion is non-trivial for general monoids. – Andrew Bacon Sep 16 '17 at 19:32

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Here are counterexamples from geometry. Lets call a set "good" for convenience when it satisfied your property.

The good set is not necessarily unique. Take $A=\Bbb R^n$ and $M$ the monoid (actually group) of translations on $\Bbb R^n$. Choosen any point $x\in \Bbb R^n$ and set $X=\{x\}$. Whereever $f$ maps $x$ to, there is a unique translation moving $x$ to $f(x)$.

A good set does not necessarily exist. Take $A=\Bbb R^n$ and $M$ the monoid (actually group) of distance preserving maps. For any set $X=\{x\}$ with only a single element, there are several distance-preserving maps in $M$ moving $x$ to $f(x)$ (with different rotations). However, when $X$ contains more than one element, then $f(x)=0$ is not distance-preserving and therefore there is no map in $M$ which can reproduce $f$ on $X$.

I initially thought that even though a good set is not unique, it might be essentially unique in the sense that all good sets $X_1$ and $X_2$ are of the same cardinality and hence (easy to prove) $X_1=iX_2$ for some invertible $i\in M$. However, as it turns out there are examples of where there can be good sets of different cardinality:

Example. There are rings $R$ for which the freely generated left $R$-modules $R^n$ and $R^m$ are isomorphic, even though $n\not=m$. This means that e.g. $R^n$ has a basis of both sizes $n$ and $m$. Choosing $A=R^n$ and $M$ the set of endomorphisms $A\to A$, any basis of $R^n$ is a good set.

M. Winter
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