Mackey's definition of group action 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?
 A: 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 $S$.
