Abstract Algebra Theory and Applications - Computer Science ...

206 CHAPTER 12 GROUP ACTIONSand the stabilizer subgroups areG 1 = G 2 = {(1), (35)(46)},G 3 = G 4 = G 5 = G 6 = {(1)}.It is easily seen that G x is a subgroup of G for each x ∈ X.Proposition 12.2 Let G be a group acting on a set X and x ∈ X. Thestabilizer group, G x , of x is a subgroup of G.Proof. Clearly, e ∈ G x since the identity fixes every element in the set X.Let g, h ∈ G x . Then gx = x and hx = x. So (gh)x = g(hx) = gx = x;hence, the product of two elements in G x is also in G x . Finally, if g ∈ G x ,then x = ex = (g −1 g)x = (g −1 )gx = g −1 x. So g −1 is in G x .□We will denote the number of elements in the fixed point set of an elementg ∈ G by |X g | and denote the number of elements in the orbit of x of x ∈ Xby |O x |. The next theorem demonstrates the relationship between orbits ofan element x ∈ X and the left cosets of G x in G.Theorem 12.3 Let G be a finite group and X a finite G-set. If x ∈ X,then |O x | = [G : G x ].Proof. We know that |G|/|G x | is the number of left cosets of G x in G byLagrange’s Theorem. We will define a bijective map φ between the orbit O xof X and the set of left cosets L Gx of G x in G. Let y ∈ O x . Then thereexists a g in G such that gx = y. Define φ by φ(y) = gG x . First we mustshow that this map is well-defined and does not depend on our selection ofg. Suppose that h is another element in G such that hx = y. Then gx = hxor x = g −1 hx; hence, g −1 h is in the stabilizer subgroup of x. Therefore,h ∈ gG x or gG x = hG x . Thus, y gets mapped to the same coset regardlessof the choice of the representative from that coset.To show that φ is one-to-one, assume that φ(x 1 ) = φ(x 2 ). Then thereexist g 1 , g 2 ∈ G such that x 1 = g 1 x and x 2 = g 2 x. Since there exists ag ∈ G x such that g 2 = g 1 g,x 2 = g 2 x = g 1 gx = g 1 x = x 1 ;consequently, the map φ is one-to-one. Finally, we must show that the mapφ is onto. Let gG x be a left coset. If gx = y, then φ(y) = gG x . □