. Lemma
In each case when a direct product acts on a set ,
we obtain both a natural action of on the set of orbits of :
and a natural action of on the set of orbits of :
Moreover the orbit of under is the set consisting of
the orbits of on that form , while the orbit of
under is the set consisting of the orbits of
on that form , and therefore the following
identity holds:
In particular each action of the form can be considered as an action of on or as an action of on .
The corresponding result on wreath products is due to W. Lehmann, and it reads as follows:
. Lemma
The following mapping is a bijection:
if is defined by In particular,
Proof: It is easy to see that is well defined. In order to prove that is injective assume , so that there exist such that But this implies
for each Therefore there must exist, for each such that Summing up there exist for which , and so
In order to show that is surjective assume that Defining in such a way that , say for all , then, for defined as above,
which gives and it completes the proof.
Exercises