@KajHansen We have the Galois group $G=\{\tau_{ij}, i=1,2,3 , j=1,2\}$, where $\tau_{ij}(\sqrt[3]{2})=\omega^{i-1}\sqrt[3]{2}, \tau_{ij}(\omega)=\omega^i$.
The subgroups $H$ of $G$ are:
#H=1: $H=\{id\}$
#H=6: $H=G$
#H=2,3
The order of the elements of $H$ is:
$ord \tau_{11}=1 \\ ord \tau_{21}=3 \\ ord \tau_{31}=3 \\ ord \tau_{12}=2 \\ ord \tau_{22}=2 \\ ord \tau_{32}=2$
$<\tau_{21}>=\{1, \tau_{21}, \tau_{21}^2\}$
Since $\tau_{31}=\tau_{21}^2, \tau_{31}^2=\tau_{21}$ we have that $ <\tau_{21}>=<\tau_{21}>$.
