It is already proved that
$1) \ \ $ $$Gal(GF(p^n)/GF(p))\equiv \mathbb Z_n$$
and $GF(p^n)$ is the splitting field of $x^{p^n}-x=x(x^{p^{n-1}}-1)$ over $GF(p)$.
Now the second one comes as
$2)\ \ $ "$F$ is a field and $E=F(\alpha )$ where $\alpha $ is a primitive $n$th root of $1$. Then $Gal(E/F)$ is isomorphic to a subgroup of $U(n)$."
But if we take a field $F$ and the splitting field, $E$, of $x^n -1$ over it then the Galois group $Gal(E/F)$ isomorphic to a subgroup of $U(n)$.
Why this happens $?$ I have gone through both the proofs and step by step they are clear but the whole picture is confusing. In the second case I can take $F=\mathbb Z_p$. Then using this result in the first one, considering the power of $x$ in the equation , I get that the Galois group is isomorphic to a subgroup of $U(p^{n-1})$.
In both the proofs, every non zero element of the basis has the for $\alpha ^k$ for some integer $k$ and the map defined $$\sigma : Gal(./.)\rightarrow \mathbb Z_n$$ is $$\sigma(\alpha)=\alpha^k$$ for some $k$. But next to this, in the second proof, it requires $k$ to be relatively prime to $n$ as $\sigma$ has to be an automorphism over $<\alpha>$ and hence comes $U(n)$ into the picture.
Then in the first proof also I could say $\sigma$ had to map $\alpha$ to $\alpha^i$ where $i$ is relatively prime to $p^{n-1}$ and relate it to a subgroup of $U(p^{n-1})$ but then again , the map $$\alpha\mapsto \alpha^p$$ is found to have order $n$ and it is cyclic.
But my question is, in the second theorem also, for any $i$ s.t. $(i,n)=1$ would not order of $$\alpha \mapsto \alpha^i$$ be $n$ $?$. Then why is this not cyclic and isomorphic to $\mathbb Z_n$ $?$
I think I am misunderstanding something seriously and this post is very clumsy, I know, apologies for that. Please explain this to me.
Thanks.
Hint: $$x^{p^n}-x=x(x^{p^n-1}-1)\neq x(x^{p^{n-1}}-1).$$