Exercise 12.1.3

Proof.

(a)

$y_1=x_1{x}_2+x_3{x}_4,y_2=(23)\cdot y_1=x_1{x}_3+x_2{x}_4,y_3=(24)\cdot y_1=x_1{x}_4+x_2{x}_3$ are distinct elements of the orbit of $y_1$.

Since $|H(y_1)|=|{\mathrm{Stab}}_{S_4}(y_1)|=8$ (see Part (b)), $|{\mathcal{}O}_{y_1}|=3$, so $y_1,y_2,y_3$ are all the elements of ${\mathcal{}O}_{y_1}$.

$${\mathcal{}O}_{y_1}=\{y_1,y_2,y_3\}=\{x_1{x}_3+x_2{x}_4,x_1{x}_3+x_2{x}_4,x_1{x}_4+x_2{x}_3\}.$$

Therefore

$$\begin{array}{llll}\hfill 𝜃(y)& =\left((y-(x_1{x}_2+x_3{x}_4)\right)\left(y-(x_1{x}_3+x_2{x}_4)\right)\left(y-(x_1{x}_4+x_2{x}_3)\right)& \hfill & \end{array}$$

Using the methods of section 2.3, we obtain with the following Sage instructions

     e = SymmetricFunctions(QQ).e()
     e1, e2, e3 , e4 =
      e([1]).expand(4),e([2]).expand(4),e([3]).expand(4), e([4]).expand(4)
     R.<y,x0,x1,x2,x3,y1,y2,y3,y4> = PolynomialRing(QQ, order = ’degrevlex’)
     J = R.ideal(e1-y1, e2-y2, e3-y3,e4-y4)
     G = J.groebner_basis()
     
     z1 = x0*x1 + x2*x3
     z2 = x0*x2 + x1*x3
     z3 = x0*x3 + x1*x2
     f = (y-(x0*x1 + x2*x3))*(y-(x0*x2 + x1*x3))*(y-(x0*x3 + x1*x2))
     
     var(’sigma_1,sigma_2,sigma_3,sigma_4’)
     g=f.reduce(G).subs(y1=sigma_1,y2=sigma_2,y3=sigma_3,y4=sigma_4)
     g.collect(y)

$$-{\sigma }_1^2{\sigma }_4-{\sigma }_2{y}^2+y^3-{\sigma }_3^2+4{\sigma }_2{\sigma }_4+\left({\sigma }_1{\sigma }_3-4{\sigma }_4\right)y.$$

So

$$𝜃(y)=y^3-{\sigma }_2{y}^2+\left({\sigma }_1{\sigma }_3-4{\sigma }_4\right)y-{\sigma }_3^2-{\sigma }_1^2{\sigma }_4+4{\sigma }_2{\sigma }_4.$$

(b)

$$(12)\cdot y_1=x_2{x}_1+x_3{x}_4=y_1,(1324)(y_1)=x_3{x}_4+x_2{x}_1=y_1,$$

therefore

$$⟨(12),(1324)⟩\subset H(y_1).$$

Moreover

$$⟨(12),(1324)⟩=\{(),(12),(1324),(13)(24),(12)(34),(14)(23),(34),(1423)\}.$$

We obtain this by hand, or with the Dimino’s algorithm, or with the Sage instructions:

     G = PermutationGroup([(1,2),(1,3,2,4)])
     G.list()

The orbit of $y_1$ contains three distinct elements $y_1,y_2,y_3$, so $|{\mathcal{}O}_{y_1}|\ge 3$. Since $|{\mathcal{}O}_{y_1}|=(S_n:H(y_1))$, $|H(y_1)|\le 8$. But $H(y_1)$ contains the 8 elements of $⟨(12),(1324)⟩$, thus

$$H(y_1)=⟨(12),(1324)⟩.$$

(c)

$(23)(1324){(23)}^{-1}=(1234)\notin H(y_1)$, so $H(y_1)$ is not normal in $S_4$.

(d)

If we number the 4 consecutive summits of the square in the order $(1,3,2,4)$, then $H(y_1)$ is isomorphic to the group generated by the rotation of angle $\pi ∕2$ corresponding to $(1324)$ and the reflection relative to the diagonal $(3,4)$ corresponding to $(12)$, and this is the dihedral group $D_8$. $$H(y_1)\simeq D_8.$$