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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 $hH_{x}$.

Then |$Ox$|= |$hH_{x}$| and |$Oy$|= |$hH_{y}$|

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

Thanks!

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    @Simonaster Conjugation is an automorphism of $G$ (and it induces one on $H$, as $H$ is normal), right? Everything should follow from that.2012-04-18

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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 $gH_xg^{-1}\subseteq G_y$).