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G is a finite group that acts transitively on $X$. And H is a normal subgroup of G. The question asks about the size of orbit under the induced action of $H$ on $X$. I pick up $x$, $y$ from set $X$ and write it as $gx=y$.

Then I established the following: $g$$H_{x}$$g^{-1}$=$H_{y}$

($x$ and $y$ may not on the same orbit of action $H$, but I still prove that they have conjugate stabilizers)

Then I try to use the Orbit-Stabilizer Theorem which indicate the bijection between the orbit of $x$ under $H$, which is $Ox$ and the left cosets of the stabilizer $h$$H_{x}$.

Then |$Ox$|= |$h$$H_{x}$| and |$Oy$|= |$h$$H_{y}$|

I feel that might be something wrong cause I just stuck here. Are there any way that I could establish $h$$H_{x}$$h^{-1}$=$H_{y}$ instead of $g$$H_{x}$$g^{-1}$=$H_{y}$?

Thanks!

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Do you need that fact? If the two stabilizers are conjugate then they have the same size and hence the same index in $H$. –  Dylan Moreland Apr 18 '12 at 2:01
    
What exactly are you stuck on? $H$ will not in general be transitive, only half-transitive (all orbits same size). –  user641 Apr 18 '12 at 2:06
    
@DylanMoreland I am not quite sure if $g$$H_{x}$$g^{-1}$=$H_{y}$ that $H_{x}$ will have the same size as $H_{y}$. And you are right, if they are the same, it's direct from the theorem that their orbits are the same size. –  Simonaster Apr 18 '12 at 2:21
    
@SteveD My stuck is that I didn't know how to use the stabilizer conjugate fact : $g$$H_{x}$$g^{-1}$=$H_{y}$ to get the result that their orbits are of the same size. –  Simonaster Apr 18 '12 at 2:23
    
@Simonaster Conjugation is an automorphism of $G$ (and it induces one on $H$, as $H$ is normal), right? Everything should follow from that. –  Dylan Moreland Apr 18 '12 at 2:25

1 Answer 1

Note that if $K\leq M\leq G$ and $g\in G$, then $[M:K]=[gMg^{-1}:gKg^{-1}]$, since conjugation by $g$ induces an automorphism of $G$. Therefore, $$|Hx| = [H:H_x] = [gHg^{-1}:gH_xg^{-1}] = [H:H_y] = |Hy|.$$ We use transitivity of $G$ to guarantee the existence of $g\in G$ with $gx=y$; we use the fact that $H$ is normal to get $gHg^{-1}=H$ and $gH_xg^{-1}=H_y$ (otherwise, we would just get $gH_xg^{-1}\subseteq G_y$).

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