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Orbits of a normal subgroup of a finite group

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3029298
#1
Jan6-10, 12:22 PM
P: 57
1. The problem statement, all variables and given/known data
If G is a finite group which acts transitively on X, and if H is a normal subgroup of G, show that the orbits of the induced action of H on X all have the same size.

3. The attempt at a solution
By the Orbit-Stabilizer theorem the size of the orbit induced by H on X is a divisor of H. This could certainly help... And H is normal, therefore H is the stabilizer of the action of conjugation. Plus the fact that points in the same orbit have conjugate stabilizers... I don't know how to put the elements together.... Can anyone hint me?
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Dick
#2
Jan6-10, 02:15 PM
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You really don't need anything fancy here. If x is an element of X then the orbit under H is Hx. If y is another element of X then there is a g in G such that y=gx (since G is transitive). So the orbit of y, Hy=Hgx. Now use that H is normal.
3029298
#3
Jan6-10, 05:41 PM
P: 57
I think I see it now... Hy=Hgx=gHg-1gx=gHx, a left coset of Hx, which has the same size as Hx, since each product of g with a member of Hx is unique and together all these products fill out gHx. The number of products is exactly as big as the size of Hx. Therefore gHx and Hx have the same size.

Correct? Thanks (again!) for your help!

Dick
#4
Jan6-10, 08:58 PM
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P: 25,228
Orbits of a normal subgroup of a finite group

Quote Quote by 3029298 View Post
I think I see it now... Hy=Hgx=gHg-1gx=gHx, a left coset of Hx, which has the same size as Hx, since each product of g with a member of Hx is unique and together all these products fill out gHx. The number of products is exactly as big as the size of Hx. Therefore gHx and Hx have the same size.

Correct? Thanks (again!) for your help!
Sure. That's correct. Not so hard, was it?
3029298
#5
Jan7-10, 06:40 AM
P: 57
no :)


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