Example. $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right)$ [22016A]

Let $F = \mathbb{Q}(\zeta_{p})$ where $\zeta_{p} = e^{ 2\pi 1 / p }$ is a primitive $p$-th root of unity.

Claim: $F / \mathbb{Q}$ is Galois.

$F$ is the splitting field of $x^{p} - 1$ over $\mathbb{Q}$ by Example 1505, so $F / \mathbb{Q}$ is normal by (4) of Proposition 2205. $\operatorname{ch}(\mathbb{Q}) = 0$, so $F / \mathbb{Q}$ is separable by (5) of Proposition 1704. Hence, $F / \mathbb{Q}$ is Galois.

Claim: $\mathrm{Gal}(F / \mathbb{Q}) \cong \left( \mathbb{Z} / p\mathbb{Z} \right)^{\times}$.

We showed that $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = \prod_{1 \leq j \leq p-1} (x - \zeta_{p}^{j})$ in Example 1505. For each integer $j$ coprime to $p$, $\zeta_{p}^{j}$ is a root of $\Phi_{p}(x)$. By Proposition 1709 and its proof, we know that $$ \mathbb{Q}(\zeta_{p}) \to \mathbb{Q}(\zeta_{p}^{j}) = \mathbb{Q}(\zeta_{p}), \qquad a(\zeta_{p}) \mapsto a(\zeta_{p}^{j}) \quad \forall a(x) \in \mathbb{Q}[x] $$ is a field isomorphism. By definition, we can check that $\sigma^{i} \circ \sigma^{j} = \sigma^{ij}$ and $\sigma^{j} = \operatorname{id}_{F}$ if and only if $j \equiv 1 \pmod{p}$. It follows that $$ \begin{align*} \left( \mathbb{Z} / p\mathbb{Z} \right)^{\times} &\to \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \\ \overline{j} &\mapsto \sigma_{j} \end{align*} $$ is well defined and is an injective group homomorphism. On the other hand, $$ \lvert \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \rvert = [\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p - 1 = \lvert \left( \mathbb{Z} / n\mathbb{Z} \right) ^{\times} \rvert . $$ Therefore, the group homomorphism is bijective and is a group isomorphism.

For general positive integer $n$, we will show that $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{n}) / \mathbb{Q} \right) \cong \left( \mathbb{Z} / n\mathbb{Z} \right)^{\times}$.