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

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 $L=K_2(\sqrt{a })$

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!

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.

• Thanks. So, since there are unique subgroups of order $4$ and $5$, then, by the Galois correspondence, there are unique subfields of orders $4$ and $5$ respectively? I am unsure what the group of units would be for $Gal(\mathbb{Q}(\zeta_{25}))/K_2$ – maths2332 Jun 2 '16 at 19:52
• Yes for the first part, as for the second one: are you asking about the "group of units" ...of a group? – DonAntonio Jun 2 '16 at 20:38
• I am just trying to figure out the second part of the question, why the extension is cyclic. I am not sure. I thought you suggested that this could be found by considering the group of units... although I can't figure out what to do – maths2332 Jun 2 '16 at 20:44
• When you refer to the group of units, you are referring to the units of the ring $\Bbb Z/p^k\Bbb Z$. The cyclicity is in the Galois group, not the group of units of the (integer ring of the) field. But the statement about cyclicity here is valid only for primes $p\ne2$. – Lubin Jun 3 '16 at 2:25

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.