Number of subfields of splitting fields of $x^5-5$ over $\mathbb{Q}$. I want to find the number of subfields of splitting fields of $x^5-5$  over $\mathbb{Q}$.

*

*By Eisenstein Criterion, $x^5-5$ is irreducible over $\mathbb{Q}.$


*Then splitting field $K$ of $x^5-5$ is Galois extension of $\mathbb{Q}$.


*Let $\zeta$ be primitive 5th root. Then Roots of $x^5-5$ consist of exactly  $\zeta \root5\of5$, $\zeta^2 \root5\of5$, $\zeta^3 \root5\of5$, $\zeta^4 \root5\of5$, $\root5\of5$.
So $$K=\mathbb{Q}(\zeta \root5\of5, \zeta^2 \root5\of5, \zeta^3 \root5\of5, \zeta^4 \root5\of5) = \mathbb{Q}(\zeta,\root5\of5)$$


*Since $K=\mathbb{Q}(\zeta)\mathbb{Q}(\root5\of5)$ and $[\mathbb{Q}(\zeta):\mathbb{Q}] = 4 \mbox{ and } [\mathbb{Q}(\root5\of5):\mathbb{Q}]=5$,
$$|Gal(K/F)|=[K:F]=5\cdot 4=20.$$
Therefore $Gal(K/F)$ is group of order 20.
So If I can find number of subgroups of $Gal(K/F)$, then I can find number of subfields of the field.
What shall I do?
 A: Since $K=\mathbb{Q}(\zeta,\sqrt[5]{5})$, an element $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ is completely determined by what $\sigma(\zeta)$ and $\sigma(\sqrt[5]{5})$ are. Note that any $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ must have 
$$\sigma(\zeta)=\zeta^k\text{ for some }1\leq k\leq 4,\qquad\quad \sigma(\sqrt[5]{5})=\zeta^r\sqrt[5]{5}\text{ for some }0\leq r\leq 4$$
(These comprise the $20$ different elements of $\mathrm{Gal}(K/\mathbb{Q})$.)
To understand $\mathrm{Gal}(K/\mathbb{Q})$, try to write it using a presentation (generators and relations). Can you think of any particular elements of $\mathrm{Gal}(K/\mathbb{Q})$ that, taken together, generate the entire group? Try to treat the two generators of the field "orthogonally" - find one automorphism that permutes the various powers of $\zeta$ and ignores $\sqrt[5]{5}$, and another automorphism that permutes the various 5th roots of $5$ and ignores any $\zeta$s (or at least, any $\zeta$s that aren't part of one of the 5th roots of $5$). Then combinations of these two should allow you to produce any desired effect on both $\zeta$ and $\sqrt[5]{5}$.
Then, determine the relations between those generating elements. That should make it easier to figure out the subgroups of $\mathrm{Gal}(K/\mathbb{Q})$ (and hence, the subfields of $K$).
A: This group of order $20$ is isomorphic to the following group of $2\times2$ matrices over the field $k=\Bbb{F}_5$
$$
G=\operatorname{Gal}(K/\Bbb{Q})\simeq\left\{
\left(\begin{array}{cc}a&b\\0&1\end{array}\right)\big\vert\,a,b\in k, a\neq0
\right\}.
$$
This isomorphism comes from mapping the automorphism $\sigma:\zeta_5\mapsto\zeta_5,
\root5\of5\mapsto\zeta_5\root5\of5$ to the matrix $\pmatrix{1&1\cr0&1\cr}$, and the
automorphisms $\tau_a:\root5\of5\mapsto\root5\of5, \zeta_5\mapsto \zeta_5^a$, $a=1,2,3,4$, to the diagonal matrices of $G$. Do check that this extends to an isomorphism. Alternatively you can bijectively map the zeros $x^5-5$ to elements of $k$, and then check that the automorphisms of $K$ are the linear bijections $x\mapsto ax+b$ from $k$ to itself.
You can then list the subgroups systematically by their orders. The Sylow $5$-subgroup consists of powers of $\sigma$, and it is a normal, hence unique subgroup of its order. The diagonal matrices of $G$ form its (cyclic) Sylow $2$-subgroup. It is equal to its own normalizer and thuse has five conjugates. This may be easier to see from the above realization of $G$ as permutations of $k$ - the Sylow $2$-subgroups are then exactly the point stabilizers.
This should give you enough structure to enable you to also list the subgroups of orders two and ten.
A: Subextensions of degree 5: $\mathbb{Q}(\sqrt[5]{5})\mid\mathbb{Q}$ is an extension of degree 5 which is not Galois (because it is contained in $\mathbb{R}$ but the minimal polynomial of $\sqrt[5]{5}$ has complex roots). Hence the corresponding subgroup of order 4 is not normal and by the Sylow theorems this means that there are 5 of them. So we have 5 subextensions of degree 5.
Subextensions of degree 4:
Again by the Sylow theorems, we know that there is only 1 subextension of degree 4 which is Galois.
Subextensions of degree 10: To see how many subgroups of order 2 there are, write the equations of the automorphisms
$$ \sigma_{jk}(\zeta)=\zeta^{j} \quad \text{and} \quad \sigma_{jk}(\sqrt[5]{5})=\zeta^{k}\sqrt[5]{5} \quad \text{for $1\leqslant j\leqslant 4$ and $0\leqslant k\leqslant 4$}$$
Count how many elements of order 2 there are: since we need $\sigma_{jk}^{2}(\zeta)=\zeta^{j\cdot j}=\zeta$, $j$ can be either 1 or 4. Similary, if $j=1$, $k$ has to be 3 and if $j=4$, $k$ has to be 0. Thus, 2 elements of order 2 and 2 subextensions of degree 10.
Subextensions of degree 2: We have an element of order 1, 2 of order 2 and 4 of order 5. Every subgroup of order 4 has exactly one element of order 2, one of order 4 and its inverse. Hence, 10 elements of order 4. This adds up to 17 elements. There are 3 missing!
Our group isn't commutative. Therefore it isn't cyclic either, and the missing elements cannot have order 20. Thus, they have order 10. They generate cyclic groups of order 10. One such cyclic group has 1 element of order 1, 4 elements of order 5 and elements of orders 2 and 10. To complete the 5 missing elements we need the 2 of order 2 and the 3 of order 10. Hence, there is only 1 cyclic subgroup of order 10.
The dihedral group of order 10 has 5 elements of order 2. We don't have so many elements of order 2. Therefore, there aren't any dihedral subgroups of order 10.
A group of order 10 is either cyclic or dihedral. Hence, there is exactly 1 subgroup of order 10 and 1 subextension of degree 2.
