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Suppose $G$ is a group and $S$ is a semigroup (a set with an associative binary operation). Just as in Groups acting on groups, say a group action of $G$ on $S$ is sensible if $g * (ss') = (g * s)(g * s')$ for all $g \in G$ and $s, s' \in S$.

What are some properties of groups acting (sensibly) on semigroups that are not always true of groups acting on sets?

Here is one idea I stumbled upon. Suppose that $F$ is a field and $G$ is a group acting on a semigroup $S$. There is a corresponding linear permutation representation of $G$ on the semigroup ring $F[S]$ ($G$ acts by permuting the basis vectors: $g * \sum_{s \in S} c_s \vec{e}_s := \sum_s c_s \vec{e}_{g * s} = \sum_s c_{g^{-1} * s} \vec{e}_s$). If $G$ acts sensibly on $S$, then the linear representation also respects multiplication: $g * (v w) = (g * v)(g * w)$ for all vectors $v, w \in F[S]$.

There is a neat consequence of this result. $S_n$ plainly acts on $[n] := \{ 1, 2, \dots, n\}$, so there is a corresponding linear permutation representation of $S_n$ on $\mathbb{R}^n$: $\sigma * (x_1, x_2, \dots, x_n) := (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, \dots, x_{\sigma^{-1}(n)})$. Of course, $S_n$ acts in the same way on $\mathbb{N}^n$ and continues to respect the addition operation of $\mathbb{N}^n$, so there is a corresponding linear permutation representation of $S_n$ on $F[x_1, x_2, \dots, x_n]$ as $\sigma * h(x_1, x_2, \dots, x_n) = h(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(n)})$. (Although it is not immediately clear in this last case, I am talking about the exact same construction of a linear representation from a group action every time I say "corresponding linear permutation representation." I think it's awesome that this works out.) Thanks to the preceding paragraph, $\sigma * (fh) = (\sigma * f)(\sigma * h)$ for all $f, h \in F[x_1, x_2, \dots, x_n]$. Thus $\sigma$ corresponds not only to an invertible linear operator on $F[x_1, x_2, \dots, x_n]$, but also a ring automorphism that fixes $F$. So by the universal property of fraction fields, $S_n$ acts linearly on $F(x_1, x_2, \dots, x_n)$ as field automorphisms fixing $F$.

jskattt797
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I think that what you are calling a "sensible action" would more usually be called an "action by semigroup automorphisms". Actually, in most contexts, when a group acts on a semigroup this property would be implicitly assumed (or else why mention the semigroup structure at all?).

Anyway, at the level of general semigroups, there is probably not much to be said. Given any set $S$ and a choice of a point $0 \in S$, defining $xy=0$ for all $x,y \in S$ makes $S$ into a semigroup. A bijection $\phi : S \to S$ will be a semigroup automorphism (i.e. will satisfy $\phi(xy)=\phi(x)\phi(y)$ for all $x,y\in S$) if and only if $\phi(0)=0$. So the semigroup automorphisms of $S$ are just the permutations of $S$ which fix the point $0$.

Obviously there is not much of a difference between general group actions on sets, and group actions on pointed sets which preserve the base point. There is a straightforward recipe for going between the two (add/remove the basepoint).

So, one response to your question "what properties do group actions on semigroups have that group actions on sets lack?" would be "not much". Without restricting attention to some special type of semigroups, basically any group action on a set can be put into this context.

Another comment: given a set $X$, you can still form a vector space $F[X]$, the vector space having the elements of $X$ as a basis. It doesn't have an interesting algebra structure, like it would in the semigroup case, but let's not worry about that. Now, if $G$ is acting on the set $X$, it also acts on the vector space $F[X]$ by linear transformations (namely ones that permute the basis). So you still get a permutation representation from a group action on a set.

Mike F
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  • Let $X^g := { x \in X : g * x = x }$ and $X^G := \cap_{g \in G} X^g$. If $X^G$ is nonempty with an element $0$, then we can trivially make $X$ into a semigroup by defining the product of everything to be $0$, and any bijection $\phi_g := [x \mapsto g * x]$ will be a semigroup automorphism since $\phi_g(0) = g * 0 = 0$, so the original action is sensible. I like this point. But how can "basically any group action on a set" become an action by semigroup automorphisms? Couldn't $X^G$ be empty? – jskattt797 May 12 '21 at 17:50
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    Did you read my third paragraph? Yes there can be actions on sets with no fixed points, my point is that there is a reversible procedure for going between (1) arbitrary group actions on sets and (2) group actions on sets with a distinguished point which fix that point. – Mike F May 12 '21 at 18:02
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    If the question is "given any group acting on a set, can a semigroup structure be put on that set so that the action becomes an action by semigroup automorphisms?" then I think the answer is "no". Working up to isomorphism, there are five different semigroups with two elements https://en.wikipedia.org/wiki/Semigroup_with_two_elements. It looks to me like for all five the only automorphism is the identity map. So, if you consider a nontrivial group action on a two element set, then it would seem that you cannot realize that action as an action on a semigroup. – Mike F May 12 '21 at 19:34
  • I'm having trouble understanding the third paragraph. Could you please explain the reversible procedure you are referencing? Here's an interesting note on the question of whether or not $X^G$ is empty: if $G$ acts transitively on $X$ and $|X| > 1$, then some $g \in G$ does not have a fixed point, so $X^G$ is empty. (Proof: if $|X^g| \geq 1$ for all $g \in G$, then $1 = \frac 1 {|G|} \left( |X| + \sum_{g \neq 1_G} |X^g| \right) \geq \frac {|X| + |G| - 1} {|G|}$ shows $|X| = 1$.) – jskattt797 May 15 '21 at 20:48
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    Given group acting on a set, you can adjoin a distinguished point to the set and extend the action by making this point fixed. Given a group acting on a set with a distinguished fixed point, you can delete the fixed point and restrict the action. This gives a bijective correspondence between (1) arbitrary group actions and (2) group actions with a distinguished fixed point. So, even though (1) and (2) are not quite the same thing, it is very easy to go between the two. Any theorem about (1) can be translated to a theorem about (2) using this correspondence, and vice versa. – Mike F May 17 '21 at 16:15
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    Yes it's true that a transitive action on a set with more than one point does not have a fixed point. I think that most people would not see that as the consequence of any calculation. Rather they would see it as a consequence of the following principles: (i) a group action on a set determines a partition of the set into "orbits", (ii) A fixed point is the same thing as an orbit with only one point in it, (iii) An action is transitive if and only if it has exactly one orbit. – Mike F May 17 '21 at 16:23
  • I guess my point is like this: Suppose mathematican A's job is to study arbitrary group actions and mathematican B's job is to study only those group actions that can be seen to preserve some semigroup structure. My point is that, for all practical purposes, these two mathematicians are working on the same topic, even though at first glance it might seem that mathematician B has a more narrow research focus than mathematician A. – Mike F May 17 '21 at 16:40