Exercise 7.3.3

Proof.

(a)

We obtain the subgroups of $D_8$ and their inclusions with the following GAP instructions:

     S:=Group((1,2,3,4),(1,3));
     T:=Group(());
     L:=IntermediateSubgroups(S,T).subgroups;
     i:=1;
     for H in L do
        Print(i, " : \t",StructureDescription(H),"\t",Order(H),"\t", H,"\t","\n");
        i:=i+1;
     od;
     Print("inclusions : \n",IntermediateSubgroups(S,T).inclusions);

We obtain:

     1 :  C2        2 Group( [ (1,3)(2,4) ] )
     2 :  C2        2 Group( [ (2,4) ] )
     3 :  C2        2 Group( [ (1,3) ] )
     4 :  C2        2 Group( [ (1,2)(3,4) ] )
     5 :  C2        2 Group( [ (1,4)(2,3) ] )
     6 :  C2 x C2   4 Group( [ (1,3)(2,4), (2,4) ] )
     7 :  C4        4 Group( [ (1,3)(2,4), (1,2,3,4) ] )
     8 :  C2 x C2   4 Group( [ (1,3)(2,4), (1,2)(3,4) ] )
     inclusions :
     [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 0, 4 ], [ 0, 5 ], [ 1, 6 ], [ 2, 6 ],
     [ 3, 6 ], [ 1, 7 ], [ 1, 8 ], [ 4, 8 ], [ 5, 8 ], [ 6, 9 ], [ 7, 9 ],
      [ 8, 9 ] ]

This corresponds to the lattice of subgroups of $G$ written in the first diagram (the node (1) corresponding to the subgroup generated by ${\sigma }^2=(1,3)(2,4)$).

We find again these results directly without computer in

$$D_8=⟨\sigma ,\tau ⟩=\{e,\sigma ,{\sigma }^2,{\sigma }^3,\tau ,\mathit{\sigma \tau },{\sigma }^2\tau ,{\sigma }^3\tau \}\simeq G,$$

where $\sigma =(1,2,3,4),\tau =(1,3)$ (cf Ex. 6.3.2(b)). Here the numbering of the roots is

$$z_1=i\sqrt[4]{2},z_2=-\sqrt[4]{2},z_3=-i\sqrt[4]{2},z_4=\sqrt[4]{2},$$

so $\tau$, which exchanges $z_1,z_3$ corresponds to the transposition $(1,3)$, and $\sigma$ to the $4$-cycle $(1,2,3,4)$.

$\sigma$ is of order 4 and generates $H=⟨\sigma ⟩=\{e,\sigma ,{\sigma }^2,{\sigma }^3\}$, $\tau$ is of order 2, and $\mathit{\sigma \tau }=\tau {\sigma }^{-1}=(1,4)(2,3)$ :

$${\sigma }^4={\tau }^2=e,\mathit{\sigma \tau }=\tau {\sigma }^{-1}.$$

Note that $\mathit{\tau \sigma }={\sigma }^{-1}\tau$ and that $\tau {\sigma }^k={\sigma }^{-k}\tau \Rightarrow \tau {\sigma }^{k+1}={\sigma }^{-k}\mathit{\tau \sigma }={\sigma }^{-k-1}\tau$. This induction proves that $\tau {\sigma }^k={\sigma }^{-k}\tau$ for all $k\in \mathbb{N}$. Moreover ${({\sigma }^k\tau )}^2={\sigma }^k\tau {\sigma }^k\tau ={\sigma }^k{\sigma }^{-k}\mathit{\tau \tau }=e$, so all the elements of the right coset $\mathit{H\tau }$ are of order 2.

We find all the subgroups of order 2 by checking the elements of order 2 in $D_8$. They are the elements of $\mathit{H\tau }=\{\tau ,\mathit{\sigma \tau },{\sigma }^2\tau ,{\sigma }^3\tau \}$, and also ${\sigma }^2\in H$: this gives all the subgroups of level 2 in the first diagram.

We know a subgroup of $G$ of order 4, the subgroup $H=⟨\sigma ⟩$.

Let $M$ be any subgroup of $G$ of order 4. If $M$ is cyclic, it is generated by an element of order 4, so $M=H=⟨\sigma ⟩=⟨{\sigma }^3⟩$.

Otherwise $M$ is isomorphic to $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, generated by to distinct elements of order 2 in $D_8\simeq G$. If one of these elements is ${\sigma }^2$, we obtain the two subgroups

$$\begin{array}{llll}\hfill H_1& =⟨{\sigma }^2,\tau ⟩=\{e,{\sigma }^2,\tau ,{\sigma }^2\tau \}=⟨{\sigma }^2,{\sigma }^2\tau ⟩& \hfill & \\ \hfill H_2& =⟨{\sigma }^2,\mathit{\sigma \tau }⟩=\{e,{\sigma }^2,\mathit{\sigma \tau },{\sigma }^3\tau \}=⟨{\sigma }^2,{\sigma }^3\tau ⟩.& \hfill & \\ \hfill & & \hfill \end{array}$$

Otherwise $M=⟨{\sigma }^k\tau ,{\sigma }^l\tau ⟩,0\le k,l\le 3,k\ne l$. As ${\sigma }^k\tau {\sigma }^l\tau ={\sigma }^{k-l}\in H$ is of order 2, ${\sigma }^{k-l}={\sigma }^2$ and so

$$M=\{e,{\sigma }^k\tau ,{\sigma }^l\tau ,{\sigma }^2\}.$$

Since $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ is generated by any pair of elements not equal to $e$, $M=⟨{\sigma }^2,{\sigma }^k\tau ⟩$, thus $M=H_1$ or $M=H_2$. We find again the subgroups of diagram 1.

(b)

We find the fixed fields $L_M$ corresponding with the subgroups $M$ of $G$. Consider the chain of fields going from $\mathbb{Q}$ to $L$ : $$\mathbb{Q}\subset \mathbb{Q}(\sqrt{2})\subset \mathbb{Q}(i\sqrt[4]{2})\subset \mathbb{Q}(i\sqrt[4]{2},\sqrt[4]{2})=\mathbb{Q}(i,\sqrt[4]{2})=L,$$

where each field is a quadratic extension of the preceding field.

Write

$$\alpha =\sqrt{2},\beta =-i\sqrt[4]{2},\gamma =\sqrt[4]{2}$$

(the symbol $\text{−}$ for $\beta$ is intended for obtaining $\sigma (\beta )=\gamma$). If we number the roots of $x^4-2$ by $x_1=\beta ,x_2=\gamma ,x_3=-\beta ,x_4=-\gamma$, the permutations corresponding to $\sigma ,\tau$ are $\tilde{\sigma }=(1,2,3,4),\tilde{\tau }=(1,3)$, with $D_8=⟨(1,2,3,4),(1,3)⟩$).

Then $(1,\alpha )$ is a basis $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$, $(1,\beta )$ a basis of $\mathbb{Q}(i\sqrt[4]{2})$ over $\mathbb{Q}(\sqrt{2})$, and $(1,\gamma )$ a basis of $\mathbb{Q}(i\sqrt[4]{2},\sqrt[4]{2})$ over $\mathbb{Q}(i\sqrt[4]{2})$, thus

$$\mathcal{}B=(1,\alpha ,\beta ,\gamma ,\mathit{\alpha \beta },\mathit{\alpha \gamma },\mathit{\beta \gamma },\mathit{\alpha \beta \gamma })$$

is a basis of $L$ over $\mathbb{Q}$.

Recall that (see Ex. 6.3.2(b))

$$\sigma (i)=i,\sigma (\sqrt[4]{2})=i\sqrt[4]{2},$$

$$\tau (i)=-i,\tau (\sqrt[4]{2})=\sqrt[4]{2}.$$

Consequently, $\sigma (\sqrt{2})={(\sigma (\sqrt[4]{2}))}^2=-\sqrt{2}$,

$$\begin{array}{llll}\hfill \sigma (\alpha )& =-\alpha ,\sigma (\beta )=\gamma ,\sigma (\gamma )=-\beta ,& \hfill & \\ \hfill \tau (\alpha )& =\alpha ,\tau (\beta )=-\beta ,\tau (\gamma )=\gamma .& \hfill & \end{array}$$

Every element $z\in L$ spans on the basis $\mathcal{}B$ under the form

$$z=a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }$$

(where $a_i\in \mathbb{Q}$)

$\text{∙}$

Computation of $L_{⟨\sigma ⟩}$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill \sigma (z)& =a_1-a_2\alpha +a_3\gamma -a_4\beta -a_5\mathit{\alpha \gamma }-a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \beta }+a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨\sigma ⟩}& ⟺0=z-\sigma (z)& \hfill & \\ \hfill & ⟺0=2a_2\alpha +(a_3+a_4)\beta +(-a_3+a_4)\gamma +(a_5-a_7)\mathit{\alpha \beta }+(a_7+a_5)\mathit{\alpha \gamma }+2a_6\mathit{\beta \gamma }& \hfill & \\ \hfill & ⟺a_2=a_3=a_4=a_5=a_6=a_7=0& \hfill & \\ \hfill & ⟺z=a_1+a_8\mathit{\alpha \beta \gamma },a_1,a_8\in \mathbb{Q}& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}[\mathit{\alpha \beta \gamma }]& \hfill & \end{array}$$ $$L_{⟨\sigma ⟩}=\mathbb{Q}(\mathit{\alpha \beta \gamma })=\mathbb{Q}(i)$$

As expected, this is a quadratic extension of $\mathbb{Q}$, corresponding with a subgroup of index 2 in $G$.

$\text{∙}$

Computation of $L_{⟨\tau ⟩}$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill \tau (z)& =a_1+a_2\alpha -a_3\beta +a_4\gamma -a_5\mathit{\alpha \beta }-a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }-a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨\tau ⟩}& ⟺0=z-\tau (z)& \hfill & \\ \hfill & ⟺0=a_3=a_5=a_6=a_8=0& \hfill & \\ \hfill & ⟺z=a_1+a_2\alpha +a_4\gamma +a_7\mathit{\alpha \gamma }(a_i\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\alpha ,\gamma )& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\gamma )& \hfill & \end{array}$$ (indeed $\alpha \in \mathbb{Q}(\gamma )$). $$L_{⟨\tau ⟩}=\mathbb{Q}(\gamma )=\mathbb{Q}(\sqrt[4]{2}).$$

$\text{∙}$

Computation of $L_{⟨{\sigma }^2⟩}$

$${\sigma }^2(\alpha )=\alpha ,{\sigma }^2(\beta )=-\beta ,{\sigma }^2(\gamma )=-\gamma .$$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill {\sigma }^2(z)& =a_1+a_2\alpha -a_3\beta -a_4\gamma -a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }-a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨{\sigma }^2⟩}& ⟺0=z-{\sigma }^2(z)& \hfill & \\ \hfill & ⟺0=a_3=a_4=a_5=a_7& \hfill & \\ \hfill & ⟺z=a_1+a_2\alpha +a_6\mathit{\beta \gamma }+a_8\mathit{\alpha \beta \gamma }(a_i\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\alpha ,\mathit{\beta \gamma })& \hfill & \\ \hfill & & \hfill \end{array}$$ $$L_{⟨{\sigma }^2⟩}=\mathbb{Q}(\alpha ,\mathit{\beta \gamma })=\mathbb{Q}(\sqrt{2},i\sqrt{2})=\mathbb{Q}(i,\sqrt{2})$$

$\text{∙}$

Computation of $L_{⟨{\sigma }^2\tau ⟩}$

$$({\sigma }^2\tau )(\alpha )=\alpha ,({\sigma }^2\tau )(\beta )=\beta ,({\sigma }^2\tau )(\gamma )=-\gamma$$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill ({\sigma }^2\tau )(z)& =a_1+a_2\alpha +a_3\beta -a_4\gamma +a_5\mathit{\alpha \beta }-a_6\mathit{\beta \gamma }-a_7\mathit{\alpha \gamma }-a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨{\sigma }^2\tau ⟩}& ⟺0=z-({\sigma }^2\tau )(z)& \hfill & \\ \hfill & ⟺a_4=a_6=a_7=a_8=0& \hfill & \\ \hfill & ⟺z=a_1+a_2\alpha +a_3\beta +a_5\mathit{\alpha \beta }(a_i\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\alpha ,\beta )& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\beta )& \hfill & \end{array}$$

$$L_{⟨{\sigma }^2\tau ⟩}=\mathbb{Q}(\beta )=\mathbb{Q}(i\sqrt[4]{2}).$$

$\text{∙}$

Computation of $L_{⟨{\sigma }^2,\tau ⟩}$ $$\begin{array}{llll}\hfill z\in L_{⟨{\sigma }^2,\tau ⟩}& ⟺z={\sigma }^2(z)\mathrm{et}z=\tau (z)& \hfill & \\ \hfill & ⟺a_3=a_4=a_5=a_7=0\mathrm{et}{a}_3=a_5=a_6=a_8=0& \hfill & \\ \hfill & ⟺a_3=a_4=a_5=a_6=a_7=a_8=0& \hfill & \\ \hfill & ⟺z=a_1+a_2\alpha ,(a_1,a_2\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\alpha )& \hfill & \end{array}$$

$$L_{⟨{\sigma }^2,\tau ⟩}=\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt{2}).$$

$\text{∙}$

Computation of $L_{⟨{\sigma }^3\tau ⟩}$

$$({\sigma }^3\tau )(\alpha )=-\alpha ,({\sigma }^3\tau )(\beta )=\gamma ,({\sigma }^3\tau )(\gamma )=\beta$$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill ({\sigma }^3\tau )(z)& =a_1-a_2\alpha +a_3\gamma +a_4\beta -a_5\mathit{\alpha \gamma }+a_6\mathit{\beta \gamma }-a_7\mathit{\alpha \beta }-a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨{\sigma }^3\tau ⟩}& ⟺0=z-({\sigma }^3\tau )(z)& \hfill & \\ \hfill & ⟺2a_2\alpha +(a_3-a_4)\beta +(a_4-a_3)\gamma +(a_5+a_7)\mathit{\alpha \beta }+(a_7+a_5)\mathit{\alpha \gamma }+2a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill & ⟺a_2=a_8=0\mathrm{et}{a}_3=a_4\mathrm{et}{a}_7=-a_5& \hfill & \\ \hfill & ⟺z=a_1+a_3(\beta +\gamma )+a_5\alpha (\beta -\gamma )+a_6\mathit{\beta \gamma }(a_i\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\beta +\gamma )& \hfill & \end{array}$$

We justify this last equivalence:

$(\beta +\gamma )[\alpha (\beta -\gamma )]=-4\in {\mathbb{Q}}^{\text{∗}}$, thus $\alpha (\beta -\gamma )\in \mathbb{Q}(\beta +\gamma )$, and ${(\beta +\gamma )}^2={\beta }^2+{\gamma }^2+2\mathit{\beta \gamma }=2\mathit{\beta \gamma }$, so $\mathit{\beta \gamma }\in \mathbb{Q}(\beta +\gamma )$.

$$L_{⟨{\sigma }^3\tau ⟩}\subset \mathbb{Q}(\beta +\gamma ).$$

Conversely $L_{⟨{\sigma }^3\tau ⟩}$ is a field (fixed field of $⟨{\sigma }^3\tau ⟩$), extension of $\mathbb{Q}$ containing $\beta +\gamma$. So it contains also $\mathbb{Q}(\beta +\gamma )$.

$$L_{⟨{\sigma }^3\tau ⟩}\supset \mathbb{Q}(\beta +\gamma ).$$

Conclusion:

$$L_{⟨{\sigma }^3\tau ⟩}=\mathbb{Q}(\beta +\gamma )=\mathbb{Q}((1-i)\sqrt[4]{2}).$$

$\text{∙}$

Computation of $L_{⟨\mathit{\sigma \tau }⟩}$

$$(\mathit{\sigma \tau })(\alpha )=-\alpha ,(\mathit{\sigma \tau })(\beta )=-\gamma ,(\mathit{\sigma \tau })(\gamma )=-\beta$$

$$\begin{array}{llll}\hfill z& =a_1+a_2\alpha +a_3\beta +a_4\gamma +a_5\mathit{\alpha \beta }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \gamma }+a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill (\tau \circ {\sigma }^3)(z)& =a_1-a_2\alpha -a_3\gamma -a_4\beta +a_5\mathit{\alpha \gamma }+a_6\mathit{\beta \gamma }+a_7\mathit{\alpha \beta }-a_8\mathit{\alpha \beta \gamma }& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill z\in L_{⟨\mathit{\sigma \tau }⟩}& ⟺0=z-(\mathit{\sigma \tau })(z)& \hfill & \\ \hfill & ⟺2a_2\alpha +(a_3+a_4)\beta +(a_4+a_3)\gamma +(a_5-a_7)\mathit{\alpha \beta }+(a_7-a_5)\mathit{\alpha \gamma }+2a_8\mathit{\alpha \beta \gamma }& \hfill & \\ \hfill & ⟺a_2=a_8=0\mathrm{et}{a}_3=-a_4\mathrm{et}{a}_7=a_5& \hfill & \\ \hfill & ⟺z=a_1+a_3(\beta -\gamma )+a_5\alpha (\beta +\gamma )+a_6\mathit{\beta \gamma }(a_i\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\beta -\gamma )& \hfill & \end{array}$$ (with a similar justification, by exchanging $\gamma$ and $-\gamma$)

Conclusion:

$$L_{⟨\mathit{\sigma \tau }⟩}=\mathbb{Q}(\beta -\gamma )=\mathbb{Q}((1-i)\sqrt[4]{2})$$

$\text{∙}$

Computation of $L_{⟨{\sigma }^2,\mathit{\sigma \tau }⟩}$ $$\begin{array}{llll}\hfill z\in L_{⟨{\sigma }^2,\mathit{\sigma \tau }⟩}& ⟺z={\sigma }^2(z)\mathrm{et}z=(\mathit{\sigma \tau })(z)& \hfill & \\ \hfill & ⟺a_3=a_4=a_5=a_7=0\mathrm{et}{a}_2=a_8=0& \hfill & \\ \hfill & ⟺z=a_1+a_6\mathit{\beta \gamma },(a_1,a_6\in \mathbb{Q})& \hfill & \\ \hfill & ⟺z\in \mathbb{Q}(\mathit{\beta \gamma })& \hfill & \end{array}$$

$$L_{⟨{\sigma }^2,\mathit{\sigma \tau }⟩}=\mathbb{Q}(\mathit{\beta \gamma })=\mathbb{Q}(i\sqrt{2}).$$

We obtain so all the fields of the second diagram.

(c)

The three subgroups of order 4 have the index 2 in $G$, therefore are normal subgroups of $G$. They correspond with three quadratic extensions of $\mathbb{Q}$, which are Galois extensions as every quadratic extension of $\mathbb{Q}$.

$\mathbb{Q}(\sqrt{2})$ is the splitting field of $x^2-2$ over $\mathbb{Q}$, $\mathbb{Q}(i)$ the splitting field of $x^2+1$, and $\mathbb{Q}(i\sqrt{2})$ the splitting field of $x^2+2$.

The subgroup $H=⟨{\sigma }^2⟩$ is normal in $G=⟨\sigma ,\tau ⟩$, since

$$\tilde{\tau }{\tilde{\sigma }}^2{\tilde{\tau }}^{-1}=(1,3)(1,3)(2,4)(1,3)=(2,4)(1,3)=(1,3)(2,4)={\tilde{\sigma }}^2,$$

thus $\tau {\sigma }^2{\tau }^{-1}={\sigma }^2\in H$ (and of course $\sigma {\sigma }^2{\sigma }^{-1}={\sigma }^2\in H$). $H$ corresponds with $\mathbb{Q}(i,\sqrt{2})$, which is so a Galois extension of $\mathbb{Q}$. $\mathbb{Q}(i,\sqrt{2})$ is the splitting field of the irreducible polynomial

$$x^4-2x^2+9=(x-i-\sqrt{2})(x-i+\sqrt{2})(x+i-\sqrt{2})(x+i+\sqrt{2})$$

(or of the reducible polynomial $(x^2-2)(x^2+1)$).

These are the only normal subgroups of $G$, as we will see in part (d).

(d)

As $\tau {\sigma }^{-1}=\mathit{\sigma \tau }$, then $\mathit{\sigma \tau }{\sigma }^{-1}={\sigma }^2\tau$, so the subgroups $⟨\tau ⟩$ and $⟨{\sigma }^2\tau ⟩$ are conjugate, thus are not normal subgroups of $G$.

Similarly ${\sigma }^3\tau ={\sigma }^{-1}\tau =\mathit{\tau \sigma }={\sigma }^{-1}(\mathit{\sigma \tau })\sigma$, so the subgroups $⟨{\sigma }^3\tau ⟩$ and $⟨\mathit{\sigma \tau }⟩$ are conjugate, and are not normal subgroups.

The subgroups $⟨\tau ⟩$ and $⟨\mathit{\sigma \tau }⟩$ are not conjugate, since $\tau$ corresponds to $(1,3)$, and $\mathit{\sigma \tau }$ to the permutation $(1,4)(2,3)$ which are not conjugate, since the conjugate of a transposition is a transposition.

So the corresponding extensions $\mathbb{Q}(\sqrt[4]{2}),\mathbb{Q}(i\sqrt[4]{2}),\mathbb{Q}((1+i)\sqrt[4]{2}),\mathbb{Q}((1-i)\sqrt[4]{2})$ are not Galois extensions of $\mathbb{Q}$.