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Let $f(x) \in F[x]$ be an irreducible separable polynomial of degree $n$ and $K|_F$ be the splitting field of $f(x)$. I want to prove the statement that if $G = \text{Gal}(K|_F)$ is cyclic then $[K:F] = n$.

So my idea is to show $|G| = n$. In general we get an injective map $\phi \colon G \to S_n$, but I am not sure of how to use the cyclic structure of $G$ to conclude $G = n$. It seems like we need to show a generator $\sigma$ of $G$ is mapped to an $n$-cycle (under $\phi$), but I don't see why this should be true.

I know that this is true for $G$ abelian, but is there another proof for cyclic $G$?

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    $\begingroup$ Hint: are cyclic groups abelian? $\endgroup$ – shoda Apr 17 '16 at 15:58
  • $\begingroup$ Yes, I know the proof for abelian $G$, but I want to avoid using this altogether. I was wondering if there is another proof using only this cyclic condition. $\endgroup$ – Kevin Sheng Apr 17 '16 at 15:59
  • $\begingroup$ But if a cyclic group is not a group with a generator, then what is it for you ? $\endgroup$ – Captain Lama Apr 17 '16 at 16:00
  • $\begingroup$ @CaptainLama I am not claiming this. $\endgroup$ – Kevin Sheng Apr 17 '16 at 16:01
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    $\begingroup$ What you need to check is : if $\sigma\in S_n$ generates a subgroup that acts transitively on $\{1,\dots,n\}$, then $\sigma$ is a $n$-cycle and the subgroup is of order $n$. And for that you may use the decomposition of $\sigma$ into cycles. $\endgroup$ – Captain Lama Apr 17 '16 at 16:06

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