# Question about subgroups of $\mathrm{Gal}(L/K)\simeq \mathbb{Z}/4\mathbb{Z}$

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?

• Note that $\sigma_2\circ\sigma_2 = \sigma_3$ and $\sigma_2\circ\sigma_3 = \sigma_4$, so $\mathrm{Gal}(L/K)=\langle \sigma_2\rangle$. Commented Jan 28, 2020 at 19:24
• So $\lbrace \sigma_1, \sigma_3 \rbrace$ is the only subgroup of order $2$, since $\sigma_3 \circ \sigma_3= \sigma_2$? Commented Jan 28, 2020 at 20:03

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 $$\sigma_2^2=\sigma_3$$.
By inspection, you might notice that there is an intermediate field $$L' = \mathbb{Q}(\sqrt{3}, i)$$ with $$[L : L'] = 2$$. Therefore, the subgroup of $$\operatorname{Gal}(L / K)$$ which fixes $$L'$$ will necessarily be a subgroup of order 2. It just remains to calculate what your elements $$\sigma_1, \sigma_2, \sigma_3, \sigma_4$$ do to $$\sqrt{3}$$, and to use this to determine which elements fix $$L'$$.
$$\Bbb Z_4$$ is cyclic and hence has only one (cyclic) subgroup of order $$2$$. So you just need to locate the element of the galois group of order $$2$$. It's $$\sigma_3$$.