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Let $L$ be a field extension of a field $K$. Suppose $L=K(a)$, and $a^n\in K$ for some integer $n$. If $L$ is Galois extension of $K$ (i.e., $L$ is the splitting field of $f(x)\in K[x]$, and $f$ is separable over $L$), then prove that the Galois group $Gal(L:K)$ is cyclic.

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This is false. Let $K=\Bbb{Q}$ and $a=e^{\pi i/8}$. Then $a^8=-1\in K$, and $L$ is the sixteenth cyclotomic field. It is Galois over the rationals, but the Galois group $\Bbb{Z}_{16}^*$ is not cyclic.

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  • $\begingroup$ OTOH, if we make the extra assumption that $K$ has a primitive $n$th root of unity, then the claim does hold. It is the easy direction in Kummer-theory. $\endgroup$ – Jyrki Lahtonen Dec 28 '13 at 8:18
  • $\begingroup$ And you can use $a=(1+i)/\sqrt2$ instead, as $\Bbb{Z}_8^*$ isn't cyclic either :-) $\endgroup$ – Jyrki Lahtonen Dec 28 '13 at 8:24
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Problem is true if we impose the condition $K$ contains a primitive root of unity.

Proof: Letting $G_n$ be group of all root unity we can show there is an injection $Gal(E/K) \to G_n$ given by $\sigma \to \frac{\sigma(a)}{a}$. So, we done.

If we impose the condition that $a^{i}$ doesn't belong to $K$ for all $i<n$ the above map would be isomorphism $\implies$ galois group has order $n$

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