I have seen many textbooks state this result without proof.
$``$ If $E$ is the splitting field for the polynomial $f=x^p-1 \in \mathbb{Q}[X]$ where $p$ is prime, then the Galois group $Gal(E:Q)\simeq \mathbb{F}_{p}^{\times} \simeq\mathbb{Z}_{p-1} . "$
I would like to know how to prove this.
The things that I know from reading are that $E=Q(\zeta)$, where $\zeta$ is a root of the irreducible polynomial $x^{p-1} + x^{p-2} \dots + 1$. So $E$ a cyclotomic extension of Q of degree $p-1$ and that the roots of $x^p - 1$ are the $p^{th}$ roots of unity. Also that the Galois group is comprised of the automorphisms of $E$ that fix $\mathbb{Q}$ and permute the roots of $f$.
As a second question, if I am asked to prove that the Galois group $Gal(E/Q)$ is cyclic, when $E$ is the splitting field for some specific $f$ and $p$ is small, say $5$ or $7$, is there an easier way to prove this than first proving the more general statement above and then drawing the conclusion from that?
Any help will be much appreciated! Thanks.
If, as you say, you already know what you wrote there, then you've already proved what you need...almost.
Let $\;\zeta=e^{2\pi i/p}\;$ , then $\;\zeta\;$ is clearly a root of $\;f(x)=x^p-1\;$ . Observe (well, prove) now that
1) All the roots of $\;f(x)\;$ are different (hint: look at $\;f'(x)\;$ )
2) $\;\forall\,k\in\Bbb Z\;,\;\;\;\left(\zeta^k\right)^p=1\;$
Deduce then that $\;[\Bbb Q(\zeta):\Bbb Q]=p-1\;$ and that $\;\Bbb Q(\zeta)\;$ is the splitting field of $\;f(x)\;$ over the rationals. Also, from (2) it follows this extension is cyclic.