The size of the orbits of a finite normal subgroup
1. The problem statement, all variables and given/known data
Let H be a finite subgroup of a group G. Verify that the formula (h,h')(x)=hxh'^{1} defines an action of H x H on G. Prove that H is a normal subgroup of G if and only if every orbit of this action contains precisely H points. 3. The attempt at a solution I solved the first part of the question: [tex]\left(\left(h,h'\right)\left(k,k'\right)\right)\left(x\right)=\left(hk, h'k'\right)\left(x\right)=hkx\left(h'k'\right)^{1}=hkxk'^{1}h'^{1}=\left(h,h'\right)\left(\left(k,k'\right)\left(x\right)\right)[/tex] This shows that the formula is a group homomorphism from H x H to G, and therefore it defines an action. But for the second part of the question I need a hint. 
Re: The size of the orbits of a finite normal subgroup
You may have omitted it intentionally because it's simple, but if not remember also to check (1,1)x = x to confirm that we have a group action.
The orbit of g is HgH. If H is normal, then [itex]HgH = g^{1}HH=g^{1}H[/itex]. In general we have [itex]gH \subseteq HgH[/itex] and [itex]Hg \subseteq HgH [/itex]. Now if [itex]HgH = H = gH = Hg[/itex] can you conclude gH=HgH=Hg? 
Re: The size of the orbits of a finite normal subgroup
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Re: The size of the orbits of a finite normal subgroup
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Re: The size of the orbits of a finite normal subgroup
Thanks!! Then I understand :)

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