Paradigmatic Examples 2

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^-1x)= H(f(g^-1x)),

for each x ÎX. Therefore there must exist, for each x, y(x) such that f'(x)= y(x)f(g^-1x). Summing up there exist ( y,g) ÎH wr _XG, for which f'=( y,g)f, and so H wr _XG(f)=H wr _XG(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 _XG(f))=G(F)=G(F'), and it completes the proof.

Paradigmatic Examples 2