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-   -   The size of the orbits of a finite normal subgroup (http://www.physicsforums.com/showthread.php?t=367745)

 3029298 Jan7-10 07:51 AM

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:
$$\left(\left(h,h'\right)\left(k,k'\right)\right)\left(x\right)=\left(hk, h'k'\right)\left(x\right)=hkx\left(h'k'\right)^{-1}=hkxk'^{-1}h'^{-1}=\left(h,h'\right)\left(\left(k,k'\right)\left(x\right)\right)$$
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.

 rasmhop Jan7-10 08:33 AM

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 $HgH = g^{-1}HH=g^{-1}H$.

In general we have $gH \subseteq HgH$ and $Hg \subseteq HgH$. Now if $|HgH| = |H| = |gH| = |Hg|$ can you conclude gH=HgH=Hg?

 3029298 Jan7-10 11:23 AM

Re: The size of the orbits of a finite normal subgroup

Quote:
 Quote by rasmhop (Post 2520607) 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 $HgH = g^{-1}HH=g^{-1}H$. In general we have $gH \subseteq HgH$ and $Hg \subseteq HgH$. Now if $|HgH| = |H| = |gH| = |Hg|$ can you conclude gH=HgH=Hg?
I understand the line of reasoning, only one point is unclear to me. You say that $HgH = g^{-1}HH=g^{-1}H$. But shouldn't this be $HgH = gHH=gH$? Because for a normal subgroup H gH=Hg for all g in G?

 rasmhop Jan7-10 11:33 AM

Re: The size of the orbits of a finite normal subgroup

Quote:
 Quote by 3029298 (Post 2520770) I understand the line of reasoning, only one point is unclear to me. You say that $HgH = g^{-1}HH=g^{-1}H$. But shouldn't this be $HgH = gHH=gH$? Because for a normal subgroup H gH=Hg for all g in G?
Yes you're right.

 3029298 Jan7-10 11:49 AM

Re: The size of the orbits of a finite normal subgroup

Thanks!! Then I understand :)

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