[proofplan] We combine two results: the injective homomorphism $\chi: \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times$ from the general cyclotomic theory (Theorem 1278), and the irreducibility of the $n$-th cyclotomic polynomial $\Phi_n$ over $\mathbb{Q}$ (Theorem 1279). Irreducibility forces $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \deg \Phi_n = \varphi(n)$, so the injection $\chi$ maps between groups of the same finite cardinality and is therefore an isomorphism. The abelianness and order of the Galois group follow immediately. [/proofplan]

[step:Construct the injective homomorphism $\chi: \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times$]Since $\operatorname{Char} \mathbb{Q} = 0$ and in particular $\operatorname{Char} \mathbb{Q} \nmid n$, the [Galois Group of Cyclotomic Extensions](/theorems/1278) (Theorem 1278) applies to the field $K = \mathbb{Q}$ with the $n$-th cyclotomic extension $L = \mathbb{Q}(\zeta_n)$. It provides an injective group homomorphism \begin{align*} \chi: \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) &\hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times, \\ \phi &\mapsto \bar{i}, \quad \text{where } \phi(\zeta_n) = \zeta_n^i. \end{align*} In particular, $|\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| \le |(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n)$.[/step]

[guided]Every automorphism $\phi \in \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is a $\mathbb{Q}$-automorphism of $\mathbb{Q}(\zeta_n)$. Since $\phi$ fixes $\mathbb{Q}$ pointwise, it must send $\zeta_n$ to another root of its minimal polynomial over $\mathbb{Q}$. In particular, $\phi(\zeta_n)$ is a root of $t^n - 1$, so $\phi(\zeta_n) = \zeta_n^i$ for some integer $i$. Moreover, $\phi(\zeta_n)$ must itself be a primitive $n$-th root of unity (since $\phi$ is an automorphism and $\zeta_n$ has multiplicative order $n$ in $\mathbb{Q}(\zeta_n)^\times$, the image $\phi(\zeta_n)$ must also have order $n$). The element $\zeta_n^i$ is a primitive $n$-th root of unity if and only if $\gcd(i, n) = 1$, i.e., $\bar{i} \in (\mathbb{Z}/n\mathbb{Z})^\times$. The [Galois Group of Cyclotomic Extensions](/theorems/1278) (Theorem 1278) verifies that the resulting map $\chi: \phi \mapsto \bar{i}$ is a well-defined injective group homomorphism. The key points are: **Well-definedness:** If $\phi(\zeta_n) = \zeta_n^i = \zeta_n^j$, then $\zeta_n^{i-j} = 1$, so $n \mid (i - j)$, so $\bar{i} = \bar{j}$ in $\mathbb{Z}/n\mathbb{Z}$. Hence the class $\bar{i}$ is uniquely determined by $\phi$. **Homomorphism property:** If $\phi(\zeta_n) = \zeta_n^i$ and $\psi(\zeta_n) = \zeta_n^j$, then $(\phi \circ \psi)(\zeta_n) = \phi(\zeta_n^j) = \phi(\zeta_n)^j = \zeta_n^{ij}$, so $\chi(\phi \circ \psi) = \overline{ij} = \bar{i} \cdot \bar{j} = \chi(\phi) \cdot \chi(\psi)$. **Injectivity:** If $\chi(\phi) = \bar{1}$, then $\phi(\zeta_n) = \zeta_n$. Since $\zeta_n$ generates $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ and $\phi$ fixes both $\mathbb{Q}$ and $\zeta_n$, the automorphism $\phi$ fixes every element of $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_n)$, so $\phi = \operatorname{id}$. Therefore $\ker(\chi) = \{\operatorname{id}\}$. The fact that $\chi$ is only an injection, not a priori a surjection, reflects that Theorem 1278 holds for any base field $K$ with $\operatorname{Char} K \nmid n$. Over a general field, $\chi$ need not be surjective — for instance, $\mathbb{F}_p(\zeta_n)/\mathbb{F}_p$ has cyclic Galois group generated by Frobenius, which is typically a proper subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$. To upgrade the injection to an isomorphism over $\mathbb{Q}$, we need a fact specific to $\mathbb{Q}$: the irreducibility of the cyclotomic polynomial.[/guided]

[step:Show $[\mathbb{Q}(\zeta_n) : \mathbb{Q}]= \varphi(n)$ using irreducibility of $\Phi_n$] By the [Irreducibility of Cyclotomic Polynomials Over $\mathbb{Q}$](/theorems/1279) (Theorem 1279), the $n$-th cyclotomic polynomial $\Phi_n \in \mathbb{Z}[t]$ is irreducible over $\mathbb{Q}$. Since $\zeta_n$ is a primitive $n$-th root of unity, $\Phi_n(\zeta_n) = 0$ by definition of $\Phi_n$ (whose roots are exactly the primitive $n$-th roots of unity). Because $\Phi_n$ is monic, lies in $\mathbb{Q}[t]$, is irreducible over $\mathbb{Q}$, and vanishes at $\zeta_n$, it follows that $\Phi_n$ is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$. By the [Structure of Simple Algebraic Extensions](/theorems/1251) (Theorem 1251), $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \deg(\operatorname{min}_\mathbb{Q}(\zeta_n)) = \deg \Phi_n = \varphi(n)$, where the last equality holds because $\Phi_n$ has exactly $\varphi(n)$ roots (the primitive $n$-th roots of unity), is monic, and has each root with multiplicity one.[/step]

[guided]The cyclotomic polynomial $\Phi_n$ is defined as \begin{align*} \Phi_n(t) := \prod_{\substack{1 \le k \le n \\ \gcd(k,n) = 1}} (t - \zeta_n^k) \in \mathbb{C}[t]. \end{align*} The product ranges over the $\varphi(n)$ integers $k$ in $\{1, \ldots, n\}$ with $\gcd(k,n) = 1$, so $\deg \Phi_n = \varphi(n)$. A standard result (proved via the identity $t^n - 1 = \prod_{d \mid n} \Phi_d(t)$ and induction on $n$) shows that $\Phi_n \in \mathbb{Z}[t]$. The [Irreducibility of Cyclotomic Polynomials Over $\mathbb{Q}$](/theorems/1279) (Theorem 1279) asserts that $\Phi_n$ is irreducible in $\mathbb{Q}[t]$. This is the deepest input to our proof. The argument (via reduction modulo primes and the Frobenius endomorphism) uses properties specific to $\mathbb{Q}$ and its ring of integers $\mathbb{Z}$ — this is why the corresponding result over general fields only gives an injection, not an isomorphism. Since $\Phi_n$ is a monic polynomial in $\mathbb{Q}[t]$ that is irreducible over $\mathbb{Q}$ and satisfies $\Phi_n(\zeta_n) = 0$, it is the unique monic irreducible polynomial in $\mathbb{Q}[t]$ vanishing at $\zeta_n$, i.e., $\Phi_n = \operatorname{min}_\mathbb{Q}(\zeta_n)$. The [Structure of Simple Algebraic Extensions](/theorems/1251) (Theorem 1251) states that for an algebraic element $\alpha$ over a field $K$, the degree $[K(\alpha) : K]$ equals $\deg(\operatorname{min}_K(\alpha))$. We verify the hypothesis: $\zeta_n$ is algebraic over $\mathbb{Q}$ (since $\Phi_n(\zeta_n) = 0$ and $\Phi_n \in \mathbb{Q}[t]$ is nonzero). Applying Theorem 1251 with $K = \mathbb{Q}$ and $\alpha = \zeta_n$: \begin{align*} [\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \deg(\operatorname{min}_\mathbb{Q}(\zeta_n)) = \deg \Phi_n = \varphi(n). \end{align*} This is the step where the proof over $\mathbb{Q}$ diverges from the general case. Over a field $K \neq \mathbb{Q}$, the minimal polynomial of $\zeta_n$ over $K$ may be a proper divisor of $\Phi_n$ (since $\Phi_n$ can factor over $K$), giving $[K(\zeta_n) : K] < \varphi(n)$ and hence $|\operatorname{Gal}(K(\zeta_n)/K)| < \varphi(n)$.[/guided]

[step:Verify that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois and compute the Galois group order]Since $\zeta_n$ is a root of $t^n - 1 \in \mathbb{Q}[t]$ and the $n$ roots $1, \zeta_n, \zeta_n^2, \ldots, \zeta_n^{n-1}$ are all powers of $\zeta_n$, every root of $t^n - 1$ lies in $\mathbb{Q}(\zeta_n)$. Therefore $\mathbb{Q}(\zeta_n)$ is a splitting field of $t^n - 1$ over $\mathbb{Q}$, so the extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is normal. Since $\operatorname{Char} \mathbb{Q} = 0$, every algebraic extension of $\mathbb{Q}$ is separable. Therefore $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is a finite Galois extension with \begin{align*} |\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| = [\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n). \end{align*}[/step]

[guided]To apply the identity $|\operatorname{Gal}(L/K)| = [L : K]$ — which holds for finite Galois extensions — we must verify that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is indeed Galois, i.e., finite, separable, and normal. **Finiteness:** $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n) < \infty$ by the previous step. **Separability:** The characteristic of $\mathbb{Q}$ is $0$. By the [Characteristic Zero Separability](/theorems/1262) theorem (Theorem 1262), every algebraic extension of a field of characteristic zero is separable. (Alternatively, the derivative of $\operatorname{min}_\mathbb{Q}(\zeta_n) = \Phi_n$ is a nonzero polynomial of degree $\varphi(n) - 1$, and $\gcd(\Phi_n, \Phi_n') = 1$ since $\Phi_n$ divides $t^n - 1$ which has no repeated roots in characteristic zero.) **Normality:** We claim $\mathbb{Q}(\zeta_n)$ is a splitting field of $t^n - 1$ over $\mathbb{Q}$. The roots of $t^n - 1$ in $\mathbb{C}$ are $1, \zeta_n, \zeta_n^2, \ldots, \zeta_n^{n-1}$, all of which are powers of $\zeta_n$ and hence lie in $\mathbb{Q}(\zeta_n)$. Moreover, $\mathbb{Q}(\zeta_n)$ is generated over $\mathbb{Q}$ by $\zeta_n$, which is itself a root of $t^n - 1$, so $\mathbb{Q}(\zeta_n)$ is the smallest field extension of $\mathbb{Q}$ containing all roots of $t^n - 1$. By the characterisation of normal extensions as splitting fields ([Normal Equals Splitting Field](/theorems/1269), Theorem 1269), $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is normal. With all three conditions verified, $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is a finite Galois extension, and the fundamental equality $|\operatorname{Gal}(L/K)| = [L : K]$ gives $|\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| = \varphi(n)$.[/guided]

[step:Conclude that $\chi$ is an isomorphism by comparing cardinalities]From the first step, $\chi: \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times$ is an injective group homomorphism. From the previous two steps, \begin{align*} |\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| = \varphi(n) = |(\mathbb{Z}/n\mathbb{Z})^\times|. \end{align*} An injective map between finite sets of the same cardinality is bijective. Therefore $\chi$ is a bijective group homomorphism, i.e., a group isomorphism: \begin{align*} \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times. \end{align*} Since $(\mathbb{Z}/n\mathbb{Z})^\times$ is an abelian group (the multiplication in $\mathbb{Z}/n\mathbb{Z}$ is commutative), the Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is abelian. Its order is $|(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n)$.[/step]

[guided]The logical structure is: we have an injective homomorphism between two finite groups. To upgrade injectivity to bijectivity, it suffices to show that the domain and codomain have the same cardinality. We have computed $|\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})| = \varphi(n)$ (using irreducibility of $\Phi_n$ and the Galois property), and $|(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n)$ by definition of the Euler totient function. Since $\chi$ is injective and both groups have $\varphi(n)$ elements, $\chi$ is surjective and hence an isomorphism. Concretely, the isomorphism says: for every $\bar{i} \in (\mathbb{Z}/n\mathbb{Z})^\times$ (i.e., every integer $i$ with $1 \le i \le n$ and $\gcd(i, n) = 1$), there exists a unique $\mathbb{Q}$-automorphism $\sigma_i \in \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ defined by $\sigma_i(\zeta_n) = \zeta_n^i$. The group operation in the Galois group corresponds to multiplication in $(\mathbb{Z}/n\mathbb{Z})^\times$: $\sigma_i \circ \sigma_j = \sigma_{ij}$ (where $ij$ is taken modulo $n$). In particular, the Galois group is abelian because $\sigma_i \circ \sigma_j = \sigma_{ij} = \sigma_{ji} = \sigma_j \circ \sigma_i$. This completes the proof that $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$, an abelian group of order $\varphi(n)$.[/guided]