Prove that the Galois group of a polynomial $p(x)=x^q-1$ is the Cyclic group of order $q-1$, where $q$ is prime.

Question

I understand why $q$ must be odd, since complex conjugation must be one of the Galios group's elements. But why must $q$ be prime?

Answer 1

The following general theorem holds for the Cyclotomic extensions:

Theorem: The splitting field over $\mathbb{Q}$ of the minimal polynomial of a $n$-nt rooth of unity is $ \mathbb{Q}(\zeta_n)$ and such extension contains all the $\zeta_k$ with $k|n$. Moreover $ \mathbb{Q}(\zeta_n)$ is a Galois extension and its Galois Group is $\left(\mathbb{Z}/n\mathbb{Z}\right)^*$

Now, in general, the polynomial $x^n-1$ is the product of the minimal polynomials of the $\zeta_k$ with $k|n$, then its splitting field is $\mathbb{Q}(\zeta_n)$.

If $n$ is prime, the group $\left(\mathbb{Z}/n\mathbb{Z}\right)^*$ is isomorphic to $\mathbb{Z}/(n-1)\mathbb{Z}$ and then is cyclic, elsewhere it may not holds.

For an example look at the Galois Group of the extension $\mathbb{Q}(\zeta_{15})$ wich is, by the theorem, isomorphic to $\mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

Furthermore you can easy check all $n$ such that the Galois Group Cyclic: $n=p^k$ with $p\neq 2$, $n=2p^k$ with $p \neq 2$, $n=2$ and $n=4$.