I am reading Mackey's book 'induced representations of groups and quantum mechanics'. I am quite baffled by his definition of group action. His definition is,
$$ (sx)y = s (xy ) , \quad \quad \forall s \in S, x \in G, y\in G \\ s e = s , \quad \text{where } e \text{ is the identity element of G}. $$
This is at odds with what I usually encounter, namely,
$$ x (y s) = (xy)s. $$
In his definition, the associative rule is broken. On the left hand side, first $x$ and then $y$, but on the right hand side, first $y$ and then $x$.
How to reconcile his definition with the more common definition?
What you seem to be used to are left group actions. A left group action of a group $G$ on a set $S$ is a multiplication map $$ G × S \to S \,, \quad (x, s) \mapsto xs $$ such that for all $x, y ∈ G$ and all $s ∈ S$, $$ x(ys) = (xy)s \,, \quad ex = x \,. $$
However, Mackey considers right group actions. A right group action of a group $G$ on a set $S$ is a multiplication map $$ S × G \to S \,, \quad (s, x) \mapsto sx $$ such that for all $x, y ∈ G$ and all $s ∈ S$, $$ (sx)y = s(xy) \,, \quad se = s \,. $$
This is not just a difference in notation: left group actions and right group actions are different things!
We can, however, translate between left group actions and right group actions: given a right group action of $G$ on $S$, denoted by “$⋅$”, we can turn it into a left group action of $G$ on $S$, denoted by “$*$”, via $$ x * s ≔ s ⋅ x^{-1} $$ for all $x ∈ G$ and all $s ∈ S$. We can use the same trick to also turn left group actions into right group actions. These two constructions are mutually inverse (since $(x^{-1})^{-1} = x$), so that we have a one-to-one correspondence between left group actions of $G$ on $S$ and right group actions of $G$ on $S$.