Compute the Galois group of \(x^n - 1 \in \QQ[x]\) as a function of \(n\).
Solution
Let \(\zeta_n=e^{2\pi i/n}\). The roots of \(x^n-1\) are exactly \[ 1,\zeta_n,\zeta_n^2,\dots,\zeta_n^{n-1}, \] so its splitting field is \(K=\QQ(\zeta_n)\).
1. Every \(\QQ\)-automorphism of \(K\) sends \(\zeta_n\) to \(\zeta_n^a\) for some \(a\in(\ZZ/n\ZZ)^\times\).
An automorphism preserves multiplicative order. Since \(\zeta_n\) has order \(n\), its image must be another primitive \(n\)-th root of unity, hence \(\zeta_n^a\) with \(\gcd(a,n)=1\).
2. The map\[ \Phi:\Gal(K/\QQ)\longrightarrow(\ZZ/n\ZZ)^\times, \qquad \sigma\longmapsto a\pmod n \]where \(\sigma(\zeta_n)=\zeta_n^a\), is an injective homomorphism.
Composition multiplies exponents modulo \(n\), so \(\Phi\) is a homomorphism. Since \(K=\QQ(\zeta_n)\), an automorphism fixing \(\zeta_n\) is the identity, hence \(\Phi\) is injective.
3. The map \(\Phi\) is surjective.
The minimal polynomial of \(\zeta_n\) over \(\QQ\) is the cyclotomic polynomial \(\Phi_n(x)\), which is irreducible of degree \(\varphi(n)\). Hence \[ |\Gal(K/\QQ)|=[K:\QQ]=\varphi(n)=|(\ZZ/n\ZZ)^\times|. \] An injective map between these finite groups is therefore bijective.
Thus \[ \boxed{\Gal(\QQ(\zeta_n)/\QQ)\cong(\ZZ/n\ZZ)^\times}. \]