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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 06:51 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:
[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.

rasmhop Jan7-10 07:33 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 [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?

3029298 Jan7-10 10:23 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 [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?

I understand the line of reasoning, only one point is unclear to me. You say that [itex]HgH = g^{-1}HH=g^{-1}H[/itex]. But shouldn't this be [itex]HgH = gHH=gH[/itex]? Because for a normal subgroup H gH=Hg for all g in G?

rasmhop Jan7-10 10:33 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 [itex]HgH = g^{-1}HH=g^{-1}H[/itex]. But shouldn't this be [itex]HgH = gHH=gH[/itex]? Because for a normal subgroup H gH=Hg for all g in G?

Yes you're right.

3029298 Jan7-10 10:49 AM

Re: The size of the orbits of a finite normal subgroup
 
Thanks!! Then I understand :)


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