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Let $p$ be a prime and $\zeta_p$ be a $p^{th}$ primitive root of unity. Let $G=\operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$, it is well known that every sub-extension of $\mathbb{Q}(\zeta_p)$ can be written as $\mathbb{Q}(\alpha_H)$, where $H\le G$ and

$$\alpha_H=\sum_{\sigma\in H}\sigma(\zeta_p).$$

If we repalce $p$ by a non-prime, $n$, can you give me an example of sub-extension that can't be written in the form $\mathbb{Q}(\alpha_H)$?

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The trick with primes is that the Galois group is cyclic, hence summing over the elements of the group produces that result just because of the group structure. In fact, since clearly $\alpha_H$ is fixed by all of $H$, the only way this can go wrong is if $\alpha_H$ has degree lower than $[G:H]$.

But this is easy, say we have $K=\Bbb Q(\zeta_n)$ with $n=12$ so that the Galois group, $G\cong V$, the Klein $4$ group. Then writing $\zeta_{12}=\zeta_4\zeta_3=i\zeta_3$ we know since $\Bbb Q(\zeta_4)\cap\Bbb Q(\zeta_3)=\Bbb Q$ that we can write a typical automorphism of $K$ as $\sigma^i\tau^j$ with $i,j\in\{0,1\}$ and $\sigma(\zeta_4)=\overline{\zeta_4}=-\zeta_4$ and fixes $\zeta_3$ and $\tau$ fixes $\zeta_4$ but take $\zeta_3\mapsto \overline{\zeta_3}$ so that their product, $\sigma\tau\stackrel{def}{=}\sigma\circ\tau$ is complex conjugation on $K$.

Now, if $H=\sigma$ we note that

$$\alpha_H=\zeta_{12}+\tau(\zeta_{12})$$

however, clearly $\tau(\zeta_{12})=\tau(\zeta_4)\tau(\zeta_3)=-\zeta_4\zeta_3$ so that $\alpha_H=0$. With this we have $K^H=\Bbb Q(\zeta_3)$ however $\Bbb Q(\alpha_H)=\Bbb Q(0)=\Bbb Q$, so the two are unequal.

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