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Given a non-abelian group G and a simply transitive left action of G on a set S, I'm defining a right action of G in the following way :

  • Identify an element $s_0$ of S to the identity of G. With the use of the left action, this defines a bijection $\chi_{s_0} : S \to G$.
  • Construct the right action of G on an element s of S as :

$ s.g = \chi^{-1}_{s_0}(\chi_{s_0}(s).g) $

Obviously this right action is non-canonical since it depends on the choice of $s_0$. On the other hand, I've read that a right action can be defined from a left action as $s.g = g^{-1}.s$. My question is therefore : are right actions always defined canonically from left actions ?

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Certainly right actions are not always defined in terms of left actions. You can easily see that the right action of a group on itself by right-multiplication is not the same as that of left-multiplication by the inverse of the acting element. Your example boils down to this difference, using the bijection $\chi_{s_0}$ to transport from $G$ to $S$.

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Perhaps it is worth emphasizing that there are two distinct issues here: the notational, versus literal multiplication-on-the-left in a group or ring.

At a notational level, to give a left action $G\times S\rightarrow S$ (transitive or not), denoted $g\times s \rightarrow g\cdot s$, is to give a map $G\times S\rightarrow S$ such that $e\cdot s=s$ for all $s\in S$, with associativity $(gh)\cdot s=g\cdot (h\cdot s)$ for all $g,h\in G$ and $s\in S$. The notational version of "right" action is identical except for putting things on the other side. At this level, given a left action $g\times s\to g\cdot s$, we can always define a right action by $s\cdot g=g^{-1}\cdot s$, where the inverse is needed for associativity.

As @Mark effectively observes, a hazard comes in over-interpreting "left" and "right" as though those adjectives necessarily applied to some internal operations in (e.g.) a group. That is, we can define a "left" action of a group $G$ on itself by right multiplication by the inverse, $g\cdot h=hg^{-1}$ (as opposed to the left action being by left multiplication, $g\cdot h=gh$). However, it would be unfortunate to misunderstand this as somehow asserting that $gh=hg^{-1}$, which is certainly not true in general.