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The size of the orbits of a finite normal subgroup

  1. Jan 7, 2010 #1
    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.
     
  2. jcsd
  3. Jan 7, 2010 #2
    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?
     
  4. Jan 7, 2010 #3
    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?
     
  5. Jan 7, 2010 #4
    Yes you're right.
     
  6. Jan 7, 2010 #5
    Thanks!! Then I understand :)
     
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