Orbits of a normal subgroup of a finite groupby 3029298 Tags: action, finite group, group theory, orbit, stabilizer 

#1
Jan610, 12:22 PM

P: 57

1. The problem statement, all variables and given/known data
If G is a finite group which acts transitively on X, and if H is a normal subgroup of G, show that the orbits of the induced action of H on X all have the same size. 3. The attempt at a solution By the OrbitStabilizer theorem the size of the orbit induced by H on X is a divisor of H. This could certainly help... And H is normal, therefore H is the stabilizer of the action of conjugation. Plus the fact that points in the same orbit have conjugate stabilizers... I don't know how to put the elements together.... Can anyone hint me? 



#2
Jan610, 02:15 PM

Sci Advisor
HW Helper
Thanks
P: 25,174

You really don't need anything fancy here. If x is an element of X then the orbit under H is Hx. If y is another element of X then there is a g in G such that y=gx (since G is transitive). So the orbit of y, Hy=Hgx. Now use that H is normal.




#3
Jan610, 05:41 PM

P: 57

I think I see it now... Hy=Hgx=gHg^{1}gx=gHx, a left coset of Hx, which has the same size as Hx, since each product of g with a member of Hx is unique and together all these products fill out gHx. The number of products is exactly as big as the size of Hx. Therefore gHx and Hx have the same size.
Correct? Thanks (again!) for your help! 



#4
Jan610, 08:58 PM

Sci Advisor
HW Helper
Thanks
P: 25,174

Orbits of a normal subgroup of a finite group 



#5
Jan710, 06:40 AM

P: 57

no :)



Register to reply 
Related Discussions  
Normal sylow subgroup of a group of order p^aq^b  Linear & Abstract Algebra  0  
normal subgroup; topological group  Differential Geometry  1  
Mobius Inversion, finite subgroup  Linear & Abstract Algebra  1  
Subgroups of a Cyclic Normal Subgroup Are Normal  Calculus & Beyond Homework  0  
Proving a nonempty finite subset of a group is a subgroup  Calculus & Beyond Homework  3 