Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.6
As in Exercise 2.2.12, let $S_4$ act on the set $R$ of all polynomials with integer coefficients in the independent variables $x_1,x_2,x_3,x_4$ by permuting the indices: $$\sigma \cdot p(x_1,x_2,x_3,x_4) = p(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)}).$$(1) Find the polynomials in the orbit of $x_1 + x_2$. (Recall from Exercise 2.2.12 that the stabilizer of this element has order 4.)
(2) Find the polynomials in the orbit of $x_1x_2 + x_3x_4$. (Recall that the stabilizer of this element has order 8.)
(3) Find the polynomials in the orbit of $(x_1 + x_2)(x_3 + x_4)$.
Solution:
(1) We have $[S_4 : \mathsf{stab}(x_1+x_2)] = 6$. A simple calculation then shows that this orbit is $$\{ x_1+x_2, x_1+x_3, x_1+x_4, x_2+x_3, x_2+x_4, x_3+x_4 \}.$$(2) We have $[S_4 : \mathsf{stab}(x_1x_2 + x_3x_4)] = 3$. A simple calculation then shows that this orbit is $$\{ x_1x_2+x_3x_4, x_1x_3+x_2x_4, x_1x_4+x_2x_3 \}.$$(3) We saw in Exercise 2.2.12, that $\mathsf{stab}((x_1+x_2)(x_3+x_4)) = \mathsf{stab}(x_1x_2+x_3x_4)$. Thus $$[S_4 : \mathsf{stab}((x_1+x_2)(x_3+x_4))] = 3.$$ A simple calculation then shows that this orbit is $$\{ (x_1+x_2)(x_3+x_4), (x_1+x_3)(x_2+x_4), (x_1+x_4)(x_2+x_3) \}.$$
