| | | 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 ´GYX can be considered as an action of H on G \\YX or as an action of G on H \\YX.
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 \\YX -> 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 \\YX | = | 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 ÎYX in such a way that f(x) ÎF'(x)=H(yx), say f(x)=yx, for all x ÎX, then, for F defined as above,
F(x)=H(f(x))=H(yx)=F'(x),which gives F(H wr XG(f))=G(F)=G(F'), and it completes the proof.
| | | Paradigmatic Examples 2 |