The size of the orbits of a finite normal subgroup

1. Jan 7, 2010

3029298

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.

2. Jan 7, 2010

rasmhop

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?

3. Jan 7, 2010

3029298

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?

4. Jan 7, 2010

rasmhop

Yes you're right.

5. Jan 7, 2010

3029298

Thanks!! Then I understand :)