Question

We consider the operation of the symmetric group S4 on the set R[x,y,z,a] through permutation of an unknown integer.
a) Calculate the length of the orbit of polynomial x2+y2+z+a. How many permutations leave this polynomial unchanged?
b) Is a polynomial of length 5 under this operation possible?
c) Show the existence of polynomials with orbit length 12 and 4.

Answer 1

Orbit Stabilizer Theorem: Suppose that G is a finite group acting on a set X. Then, for any x in X , |Orbit(x)| = [G : Stabilizer(x)]

Up to isomorphism, we can consider S₄ to be equal to the symmetric group on {x,y,z,a}.

S₄ operates on R[x,y,z,a] by the group operation .f(x,y,r,a)=f(x,y,z,a) for all in S₄ and f in R[x,y,z,a].

a)

Let us compute the cardinality of the stabilizer of the polynomial g(x,y,z,a) = x² + y² + z + a.

Observe that if is in the stabilizer of g in S₄, then .g = g and thus, g(x,y,z,a) = g(x,y,z,a). This implies that (x)² + (y)² + z + a = x² + y² + z + a. Therefore, {x,y} = {x,y} and {z,a} = {z,a} as sets. Thus, |_{x,y} is a permutation of {x,y} and |_{z,a} is a permutation of {z,a}. Thus, the total number of such is 2.2=4 (the set {x,y} has two permutations and corresponding to each of these permutations, the set {z,a} has two permutations). Intact, the stabilizer of g is { identity, (x y), (z a), (x y)(z a) }.

Thus, this polynomial g(x,y,z,a)= x² + y² + z + a is left unchanged by 4 permutations.

By the orbit stabilizer theorem, the number of elements in the orbit of the polynomial is the index of the stabilizer in S₄. Since the stabilizer of g has 4 elements, hence the length of the orbit of the polynomial is |S₄|/4 = 24/4 = 6

b)

Suppose that there is a polynomial h whose orbit is of length 5. Then, the orbit stabilizer theorem states that the cardinality of the orbit of h equal to the index of its stabilizer in S₄. In other words, |Orbit(h)| = |S₄| / |Stabilizer(h)| and thus, 5 = 24 / Stabilizer (h). This implies that 5 | 24 which is a contradiction.

Hence, there is no polynomial whose orbit is of length 5, under this action.

c)

Consider the polynomial F(x,y,z,a) = x² + y² + z³ + a. If is in the stabilizer of F, then .F = F implies that (x)² + (y)² + (z)³ + (a) = x² + y² + z³ + a and thus, |_{x,y} is a permutation of {x,y} and fixes z and a. Thus, | Stabilizer(F) | = 2 (corresponding to the two permutations of {x,y} which keep z and a fixed). By the orbit stabilizer theorem, the orbit of F is of length 24/2 = 12.

Again consider the polynomial G(x,y,z,a) = x + y + z + a². If is in the stabilizer of G, then .G = G implies that x + ​​​​​​y + z + (a)² = x + y + z + a². Thus, as before, |_{x,y,z} is a permutation of {x,y,z} and fixes a. Thus, the number of such is 3!=6 (corresponding to the 6 permutations of {x,y,z} which fix a). Thus, |Stabilizer(G)| = 6. By the orbit stabilizer theorem, the length of the orbit of G is 24/6 = 4.