Exercise 7.2.6

Proof. $⟨\sigma ⟩,⟨\tau ⟩,⟨\mathit{\sigma \tau }⟩,⟨{\sigma }^2\tau ⟩,\{e\},G$ are subgroups of $G=\mathrm{Gal}(\mathbb{Q}(\omega ,\sqrt[3]{2})∕\mathbb{Q})\simeq S_3$, corresponding to the subgroups of $S_3$ given by $⟨(1,2,3)⟩,⟨(1,2)⟩,⟨(2,3)⟩,⟨(1,3)⟩,\{()\},S_3$. We show that $S_3$ has no other subgroup.

The order of a subgroup $H$ of $S_3$ divides 6. If $|H|=1,H=\{()\}$, if $|H|=6,H=S_3$. If $|H|=3$, $H$ is cyclic of order 3. As the only elements of order 3 of $S_3$ are $\tilde{\sigma }=(1,2,3)$ and $(1,3,2)={\tilde{\sigma }}^{-1}$, $H=⟨\tilde{\sigma }⟩$.

If $|H|=2$, is cyclic of order 2. The only elements of $S_3$ of order 2 are the three transpositions $(1,2),(2,3),(1,3)$. $S_3$, so $H\in \{⟨(1,2)⟩,⟨(2,3)⟩,⟨(1,3)⟩\}$. $S_3$ has exactly 6 subgroups, therefore $\mathrm{Gal}(\mathbb{Q}(\omega ,\sqrt[3]{2})∕\mathbb{Q})\simeq S_3$ has exactly six subgroups given in diagram (7.7). □