### Paradigmatic Examples 2

**Lemma: **
*
In each case when a direct product **H ´G* acts on a set *M*,
we obtain both a natural action of *H* on the set of orbits of *G*:
*H ´(G \\M) -> G \\M :(h,G(m)) -> G(hm),
*

and a natural action of *G* on the set of orbits of *H*:
*G ´(H \\M) -> H \\M :(g,H(m)) -> H(gm).
*

Moreover the orbit of *G(m) ÎG \\M* under *H* is the set consisting of
the orbits of *G* on *M* that form *(H ´G)(m)*, while the orbit of
*H(m) ÎH \\M* under *G* is the set consisting of the orbits of
*H* on *M* that form *(H ´G)(m)*, and therefore the following
identity holds:
* | H \\(G \\M) | = | G \\(H \\M) | = | (H ´G)
\\M
| .
*

In particular each action of the form
_{H ´G}Y^{X} can be considered as an action of *H* on *G \\Y*^{X} or
as an action of *G* on *H \\Y*^{X}.

The corresponding result on wreath products is due to W. Lehmann, and it
reads as follows:

**Lemma: **
*
The following mapping is a bijection:
** F:H wr *_{X} G \\Y^{X} -> G \\((H \\Y)^{X} ) :H wr _{X} G (f) -> G(F),

if *F Î(H \\Y)*^{X} is defined by *F(x):=H(f(x)).* In particular,
* | H wr *_{X} G \\Y^{X} | = | G \\(H \\Y)^{X} ) | .

Proof: It is easy to see that * F* is well defined.
In order to prove that * F* is injective assume
*G(F)=G(F')*, so that there exist
*g ÎG* such that *F'=F o g*^{-1}. But this implies

*H(f'(x))=
F'(x)=F(g*^{-1}x)=
H(f(g^{-1}x)),

for each *x ÎX.* Therefore there must exist, for each *x,*
* y(x)* such that *f'(x)= y(x)f(g*^{-1}x). Summing up there exist
*( y,g) ÎH wr *_{X}G, for which *f'=( y,g)f*, and so
*H wr *_{X}G(f)=H wr _{X}G(f').
In order to show that * F* is surjective assume that
*G(F')
ÎG \\(H \\Y)*^{X}.
Defining *f ÎY*^{X} in such a way that
*f(x) ÎF'(x)=H(y*_{x}), say *f(x)=y*_{x}, for all *x ÎX*, then, for *F*
defined as above,

*F(x)=H(f(x))=H(y*_{x})=F'(x),

which gives
* F(H wr *_{X}G(f))=G(F)=G(F'), and it completes the proof.

harald.fripertinger@kfunigraz.ac.at,

last changed: August 28, 2001