Let $L=\mathbb{Q}(\sqrt[4]{3},i)$ and $K=\mathbb{Q}(i)$.
I know the Galois group is $\mathrm{Gal}(L/K)\simeq \mathbb{Z}/4\mathbb{Z}$.
I have a question about the subgroups of $\mathbb{Z}/4\mathbb{Z}$.
How to find the subgroups of $\mathrm{Gal}(L/K)\simeq \mathbb{Z}/4\mathbb{Z}$? And which is the only subgroup of order $2$?
I tried:
$\mathrm{Gal}(L/K)$ acts transitively on the roots of $f(x)=x^4-3$, so there exist $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ with $\sigma_1(\sqrt[4]{3})=\sqrt[4]{3}, \sigma_2(\sqrt[4]{3})=i\sqrt[4]{3}, \sigma_3(\sqrt[4]{3})=-\sqrt[4]{3}$ and $\sigma_4(\sqrt[4]{3})=-i\sqrt[4]{3}$.
I know that the only subgroup of order $2$ is $\lbrace \sigma_1, \sigma_3 \rbrace$. But how can it be determined?
How to conclude now?
Best Answer
It is known that the subgroups of a cyclic group $G$ are cyclic. If $\sigma$ is a generator of the group $G$, a subgroup $H$ of $G$ is generated by some $\sigma^k$. Further more, if $G$ has order $n$, the order of $H$, i.e. the order of $\sigma^k$ is $$\bigl|\langle\,\sigma^k\,\rangle\bigr|=\frac{\bigl|\langle\,\sigma\,\rangle\bigr|}{\gcd\bigl(k,\bigl|\langle\,\sigma\,\rangle\bigr|\bigr)}.$$ Therefore, in the present case, a proper subgroup has order $1$ or $2$. A generator is $\sigma_2$. The subgroup of order $1$ is generated by the identity ($\sigma_1$ with your notations), and the subgroup of order $2$ is generated by $\sigma_2^2=\sigma_3$.