Let $S$ be a semigroup and $X$ any set.

Define a left action of $S$ on $X$ to be a map $\sigma: S \times X \rightarrow X$ with the property that $(st)x = s(tx)$, where we define $gx = \sigma(g,x)$ for all $g \in G$, $x \in X$.

Similarly, a right action of $S$ on $X$ is a map $\tau: X \times S \rightarrow X$ with the property $x(st) = (xs)t$, where $xg = \tau(x,g)$ for all $g \in G$, $x \in X$.

Is there a natural bijection between left and right actions of $S$? Does a left action induce in some natural way a right action?

Basically, if I have proven something for every left action of a particular semigroup, is there a shortcut to prove it for every right action as well?

  • $\begingroup$ Just to be clear, why is the definition of right(left) action as above well-defined? Thanks $\endgroup$ – mrs Apr 3 '13 at 18:01
  • $\begingroup$ @BabakS.: What exactly would it mean for it to be not well-defined? $\endgroup$ – Tara B Apr 3 '13 at 18:16
  • $\begingroup$ @TaraB: I was confused about any in "..$X$ any set" above. It is more clear by your explanation. In fact, I didn't see this concept in Howie. $\endgroup$ – mrs Apr 3 '13 at 18:18
  • $\begingroup$ @BabakS.: Ah, I see. 'Any' is quite a tricky word, which people should be a bit more careful with. The usage here is completely fine though. $\endgroup$ – Tara B Apr 3 '13 at 18:20
  • $\begingroup$ @spin: What would be an example of 'something' in your final sentence? As you can see, it might somewhat depend on the 'something'. $\endgroup$ – Tara B Apr 3 '13 at 18:21

No. Right actions and left actions of a semigroup can be significantly different.
(I was about to give the same example as Boris.)

It's actually very common for the left action of a semigroup on itself to be different from the right action. This is why we have the concept of left and right Cayley graphs for semigroups.

What you do get is that a left action of a semigroup $S$ induces a right action of the 'opposite semigroup' of $S$, which I'll denote $S^{\operatorname{opp}}$. This is the semigroup with the same underlying set as $S$, but with multiplication reversed, so that in $S^{\operatorname{opp}}$, for any $s,t\in S$ we have $st = t\circ s$, where $\circ$ is the binary operation of $S$.

In general a semigroup is not necessarily isomorphic to its opposite semigroup, but it is what is called anti-isomorphic. An anti-isomorphism is a bijective homomorphism $\phi: S\rightarrow T$ such that $\phi(xy) = \phi(y)\phi(x)$ for all $x,y\in S$.
Note that the reason you never hear of anti-isomorphisms in group theory is that anti-isomorphic groups are isomorphic. If $G$ and $H$ are groups and $\phi: G\to H$ is an anti-isomorphism, then $\psi: G\to H$ given by $x\mapsto [\phi(x)]^{-1}$ is an isomorphism.

If you want to prove a result about actions of some class of semigroups, then if the class is closed under anti-isomorphism (which many natural classes of semigroups are), it will suffice to only prove the result for left actions.

  • $\begingroup$ Thanks, $S^{\operatorname{opp}}$ was just what I needed. There was this theorem in a book that concerns the continuity of a left action for a semigroup with a topology, so with $S^{\operatorname{opp}}$ I was able to prove its analogue for right actions. $\endgroup$ – spin Apr 3 '13 at 18:24
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    $\begingroup$ Ah, great! Yes, for general results like that $S^{\operatorname{opp}}$ is exactly what you need. Glad to have helped! $\endgroup$ – Tara B Apr 3 '13 at 18:27

Left and right actions of semigroups (unlike groups!) can be arranged quite differently. Let $S$ be a semigroups of left zeros (i.e. with identity $ab=a$). Consider its actions on itself by multiplication (i.e. $X=S$). Then for the right action $X$ consists of fixed points, but for left action every $a\in S$ maps all $X$ in the single point $a$.

  • $\begingroup$ I was just typing up the same example! =] $\endgroup$ – Tara B Apr 3 '13 at 18:02
  • $\begingroup$ Oh yes, which means +1, because obviously I thought it was a good idea. $\endgroup$ – Tara B Apr 3 '13 at 18:15
  • $\begingroup$ @Tara B: "I was just typing up the same example!" - Of course, this is the first thing that comes to mind. Concerning your answer, it is worth to note (for OP) that anti-isomorphic groups are isomorphic. $\endgroup$ – Boris Novikov Apr 3 '13 at 18:44
  • $\begingroup$ Ah yes, I was going to put that in, and then I either forgot or wanted to keep it short. But you're probably right that I should mention it. $\endgroup$ – Tara B Apr 3 '13 at 18:57

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