Existence of subfields such that $[\mathbb{Q}(\zeta_{25}) : K_1]=4$ and $[\mathbb{Q}(\zeta_{25}) : K_2]=5$

Question

I have the following two questions questions I am working on and am a little stuck :

Let $L=\mathbb{Q}(\zeta_{25})$ where $\zeta_{25}$ is the primitive $n$-th root of unity.

Prove that there are unique subfields $K_1$ and $K_2$ in $L$ such that $[L:K_1]=4$ and $[L:K_2]=5$

I am not sure if this is related but $\mathbb{\phi(25)=20} \implies$ the order of $Gal(\mathbb{Q}(\zeta_{25})/\mathbb{Q}) \simeq Z_{25 }^*=Z_{20}$ (I think)

Can we say that all there exists subfields of orders equal to the divisor of our Galois group? i.e. orders $1, 2, 4, 5, 10, 20$?

Prove that $L/K_2$ is cyclic and find $a \in K_2$ such that $L=K_2(\sqrt[5]{a })$

Need to show that $Gal(L/K_2)=Z_n$ for some $n\in \mathbb{Z}$ - is that right?

$Gal(L/K_2)=Gal(\mathbb{Q}(\zeta_{25})/K_2)$

We know that $[\mathbb{Q}(\zeta_{25}) : K_2]=5$ - is it true that the only groups of order $5$ is $Z_5$ respectively (I am not convinced)

If so what would $a$ be - some primitive root of unity, $-5$?

I hope you can help me with this as I need to improve my uunderstanding of these ideas (degres of field extensions and Galois groups) - thanks!

Answer 1

Remember:

1) A finite cyclic group of order $\;n\;$ has one unique subgroup of order $\;d\;$ for any $\;d\,\mid\,n\;$ ;

2) The group of units modulo $\;p^k\;,\;\;p\;$ a prime, is cyclic.

Answer 2

It really helps to know what the minimal polynomial for $\zeta_{25}$ is: it’s $\Phi_{25}=X^{20}+X^{15}+X^{10}+X^5+1$. This makes sense, since the minimal polynomial fo $\zeta_5$ is $\Phi_5(X)=X^4+X^3+X^2+X+1$, and clearly any fifth root of a primitive fifth root of unity will be a primitive $25$-th root of unity.

For the rest of your question, I think Joanpemo’s answer does most of the work, except perhaps for the remarks that (1) up to isomorphism there is only one cyclic group of each order; and (2) every subgroup and every quotient group of a cyclic group is cyclic.