[proofplan] The proof establishes three properties of the map $\chi$ in sequence. First, well-definedness: since $\zeta$ is a primitive $n$-th root of unity, its image $\phi(\zeta)$ under any $K$-automorphism $\phi$ is again a primitive $n$-th root of unity (because $\phi$ preserves the order), so $\phi(\zeta) = \zeta^i$ for a unique $\bar{i} \in (\mathbb{Z}/n\mathbb{Z})^\times$. Second, the homomorphism property follows from composing automorphisms: $(\phi \circ \psi)(\zeta) = \phi(\zeta^j) = \zeta^{ij}$. Third, injectivity holds because $\zeta$ generates $L$ over $K$, so any automorphism fixing $\zeta$ is the identity. The consequences — abelianness and the divisibility $[L:K] \mid \varphi(n)$ — are immediate from the injection into the abelian group $(\mathbb{Z}/n\mathbb{Z})^\times$. [/proofplan] [step:Show that $\phi(\zeta)$ is a primitive $n$-th root of unity for every $\phi \in \operatorname{Gal}(L/K)$] Let $\phi \in \operatorname{Gal}(L/K)$. Since $\zeta$ is a primitive $n$-th root of unity in $L$, we have $\zeta^n = 1$ and $\zeta^k \neq 1$ for all $1 \le k < n$. Applying $\phi$ to the equation $\zeta^n = 1$ and using that $\phi$ is a ring homomorphism fixing $1 \in K$: \begin{align*} \phi(\zeta)^n = \phi(\zeta^n) = \phi(1) = 1. \end{align*} Hence $\phi(\zeta)$ is an $n$-th root of unity. It remains to show that $\phi(\zeta)$ is primitive, i.e., that $\operatorname{ord}(\phi(\zeta)) = n$. Suppose for contradiction that $\phi(\zeta)^k = 1$ for some $1 \le k < n$. Then $\phi(\zeta^k) = \phi(\zeta)^k = 1 = \phi(1)$. Since $\phi$ is injective (as a field homomorphism), this forces $\zeta^k = 1$, contradicting the primitivity of $\zeta$. Therefore $\phi(\zeta)$ has exact order $n$. [guided] The goal of this step is to show that every $K$-automorphism of $L$ sends $\zeta$ to another primitive $n$-th root of unity — not merely an $n$-th root of unity. This distinction is essential: without primitivity, the exponent $i$ in $\phi(\zeta) = \zeta^i$ could satisfy $\gcd(i, n) > 1$, and the target of $\chi$ would be $\mathbb{Z}/n\mathbb{Z}$ rather than the unit group $(\mathbb{Z}/n\mathbb{Z})^\times$. Let $\phi \in \operatorname{Gal}(L/K)$. Since $\phi$ is a ring homomorphism $L \to L$ fixing $K$ pointwise, and $1 \in K$, we have $\phi(1) = 1$. The defining property $\zeta^n = 1$ gives: \begin{align*} \phi(\zeta)^n = \phi(\zeta^n) = \phi(1) = 1, \end{align*} using that $\phi$ respects multiplication: $\phi(a \cdot b) = \phi(a) \cdot \phi(b)$ for all $a, b \in L$. This shows $\phi(\zeta)$ is an $n$-th root of unity. To upgrade from "$n$-th root" to "primitive $n$-th root," we use injectivity. Suppose $\phi(\zeta)$ had order $k < n$. Then $\phi(\zeta)^k = 1$, which means $\phi(\zeta^k) = 1 = \phi(1)$. Every field homomorphism is injective (its kernel is an ideal of a field, hence $\{0\}$), so $\phi(\zeta^k) = \phi(1)$ implies $\zeta^k = 1$. But $\zeta$ has order $n > k$, a contradiction. Therefore $\phi(\zeta)$ has exact order $n$ in $L^\times$, i.e., $\phi(\zeta)$ is a primitive $n$-th root of unity. Since the $n$-th roots of unity in $L$ form the cyclic group $\langle \zeta \rangle = \{1, \zeta, \zeta^2, \ldots, \zeta^{n-1}\}$, and the primitive $n$-th roots of unity are exactly the generators of this group, we can write $\phi(\zeta) = \zeta^i$ for some $i$ with $\gcd(i, n) = 1$, i.e., $\bar{i} \in (\mathbb{Z}/n\mathbb{Z})^\times$. [/guided] [/step] [step:Verify that $\chi$ is well-defined as a map into $(\mathbb{Z}/n\mathbb{Z})^\times$] By the previous step, for each $\phi \in \operatorname{Gal}(L/K)$, the image $\phi(\zeta)$ is a primitive $n$-th root of unity. Since the primitive $n$-th roots of unity are exactly $\{\zeta^i : \gcd(i, n) = 1\}$, there exists $i \in \{0, 1, \ldots, n-1\}$ with $\gcd(i, n) = 1$ such that $\phi(\zeta) = \zeta^i$. We must verify that the residue class $\bar{i} \in \mathbb{Z}/n\mathbb{Z}$ is uniquely determined by $\phi$. Suppose $\phi(\zeta) = \zeta^i = \zeta^j$ for integers $i$ and $j$. Then $\zeta^{i-j} = 1$, so $n \mid (i - j)$ (since $\zeta$ has exact order $n$), which means $\bar{i} = \bar{j}$ in $\mathbb{Z}/n\mathbb{Z}$. Therefore $\chi(\phi) := \bar{i}$ is well-defined, and $\chi(\phi) \in (\mathbb{Z}/n\mathbb{Z})^\times$ since $\gcd(i, n) = 1$. [guided] Two things must be checked for well-definedness: (1) that the integer $i$ is determined modulo $n$ (so the map lands in $\mathbb{Z}/n\mathbb{Z}$), and (2) that $\bar{i}$ is a unit in $\mathbb{Z}/n\mathbb{Z}$ (so the map lands in the correct target $(\mathbb{Z}/n\mathbb{Z})^\times$). For (1): the group of $n$-th roots of unity in $L$ is the cyclic group $\langle \zeta \rangle$ of order $n$. In a cyclic group of order $n$, the equation $\zeta^i = \zeta^j$ holds if and only if $n \mid (i - j)$. (This is because $\zeta^{i-j} = 1$ and $\operatorname{ord}(\zeta) = n$, so $n \mid (i-j)$.) Therefore the exponent $i$ is uniquely determined as an element of $\mathbb{Z}/n\mathbb{Z}$. For (2): the previous step showed that $\phi(\zeta)$ has order $n$. The element $\zeta^i$ has order $n / \gcd(i, n)$ in the cyclic group $\langle \zeta \rangle$. For this to equal $n$, we need $\gcd(i, n) = 1$, which is precisely the condition $\bar{i} \in (\mathbb{Z}/n\mathbb{Z})^\times$. Hence $\chi: \operatorname{Gal}(L/K) \to (\mathbb{Z}/n\mathbb{Z})^\times$, $\phi \mapsto \bar{i}$ where $\phi(\zeta) = \zeta^i$, is a well-defined function. [/guided] [/step] [step:Show that $\chi$ is a group homomorphism] Let $\phi, \psi \in \operatorname{Gal}(L/K)$ with $\chi(\phi) = \bar{i}$ and $\chi(\psi) = \bar{j}$, so that $\phi(\zeta) = \zeta^i$ and $\psi(\zeta) = \zeta^j$. We compute: \begin{align*} (\phi \circ \psi)(\zeta) &= \phi(\psi(\zeta)) = \phi(\zeta^j) = \phi(\zeta)^j = (\zeta^i)^j = \zeta^{ij}. \end{align*} The third equality uses that $\phi$ is a ring homomorphism: $\phi(\zeta^j) = \phi(\zeta \cdot \zeta \cdots \zeta) = \phi(\zeta)^j$. Therefore $\chi(\phi \circ \psi) = \overline{ij} = \bar{i} \cdot \bar{j} = \chi(\phi) \cdot \chi(\psi)$. Since the group operation in $\operatorname{Gal}(L/K)$ is composition and the group operation in $(\mathbb{Z}/n\mathbb{Z})^\times$ is multiplication, this shows $\chi$ is a group homomorphism. [guided] The key computation is expanding $(\phi \circ \psi)(\zeta)$. The group operation in $\operatorname{Gal}(L/K)$ is composition of automorphisms, while the group operation in $(\mathbb{Z}/n\mathbb{Z})^\times$ is multiplication of residue classes. We must show that $\chi$ intertwines these two operations. Write $\phi(\zeta) = \zeta^i$ and $\psi(\zeta) = \zeta^j$. Then: \begin{align*} (\phi \circ \psi)(\zeta) = \phi(\psi(\zeta)) = \phi(\zeta^j). \end{align*} Now, $\zeta^j$ is the $j$-fold product $\zeta \cdot \zeta \cdots \zeta$. Since $\phi$ is a ring homomorphism (in particular, it respects multiplication), we have: \begin{align*} \phi(\zeta^j) = \phi(\underbrace{\zeta \cdot \zeta \cdots \zeta}_{j}) = \underbrace{\phi(\zeta) \cdot \phi(\zeta) \cdots \phi(\zeta)}_{j} = \phi(\zeta)^j = (\zeta^i)^j = \zeta^{ij}. \end{align*} Therefore $\chi(\phi \circ \psi) = \overline{ij}$. On the other hand, $\chi(\phi) \cdot \chi(\psi) = \bar{i} \cdot \bar{j} = \overline{ij}$ in $(\mathbb{Z}/n\mathbb{Z})^\times$ (multiplication of residue classes is defined by $\bar{i} \cdot \bar{j} = \overline{ij}$). Hence $\chi(\phi \circ \psi) = \chi(\phi) \cdot \chi(\psi)$. Note that the argument uses nothing about $K$ or the specific extension $L/K$ — only that $\phi$ and $\psi$ are ring homomorphisms and $\zeta$ generates the group of $n$-th roots of unity. This generality reflects the fact that the homomorphism $\chi$ is a purely group-theoretic consequence of acting on a cyclic group by automorphisms. [/guided] [/step] [step:Prove injectivity of $\chi$] Suppose $\phi \in \ker(\chi)$, i.e., $\chi(\phi) = \bar{1}$. Then $\phi(\zeta) = \zeta^1 = \zeta$. Since $L = K(\zeta)$, every element of $L$ is a $K$-linear combination of powers of $\zeta$: each $\alpha \in L$ can be written as \begin{align*} \alpha = a_0 + a_1 \zeta + a_2 \zeta^2 + \cdots + a_{m-1} \zeta^{m-1} \end{align*} for some $m \le [L:K]$ and coefficients $a_0, \ldots, a_{m-1} \in K$. (Precisely, $\{1, \zeta, \ldots, \zeta^{m-1}\}$ spans $L$ over $K$ where $m = [L:K]$.) Since $\phi$ fixes $K$ pointwise and $\phi(\zeta) = \zeta$: \begin{align*} \phi(\alpha) &= \phi(a_0 + a_1 \zeta + \cdots + a_{m-1} \zeta^{m-1}) \\ &= a_0 + a_1 \phi(\zeta) + \cdots + a_{m-1} \phi(\zeta)^{m-1} \\ &= a_0 + a_1 \zeta + \cdots + a_{m-1} \zeta^{m-1} = \alpha. \end{align*} The second equality uses that $\phi$ is a $K$-linear ring homomorphism: $\phi(a_k) = a_k$ for $a_k \in K$, and $\phi(\zeta^k) = \phi(\zeta)^k$. Since $\phi(\alpha) = \alpha$ for all $\alpha \in L$, we have $\phi = \operatorname{id}_L$. Therefore $\ker(\chi) = \{\operatorname{id}_L\}$, and $\chi$ is injective. [guided] The injectivity argument rests on the fact that $\zeta$ **generates** $L$ over $K$. A $K$-automorphism of $L$ is determined by where it sends the generators of $L$ over $K$. Since $L = K(\zeta)$, the single element $\zeta$ generates $L$ as a $K$-algebra, so knowing $\phi(\zeta)$ determines $\phi$ completely. More precisely, suppose $\phi \in \ker(\chi)$, meaning $\chi(\phi) = \bar{1}$, i.e., $\phi(\zeta) = \zeta$. We must show $\phi = \operatorname{id}_L$. Every element $\alpha \in L = K(\zeta)$ can be expressed as a polynomial in $\zeta$ with coefficients in $K$. Let $f = \operatorname{min}_K(\zeta) \in K[t]$ be the minimal polynomial of $\zeta$ over $K$, with $\deg f = [L : K] =: m$. Then $\{1, \zeta, \zeta^2, \ldots, \zeta^{m-1}\}$ is a $K$-basis for $L$, and every $\alpha \in L$ has a unique representation: \begin{align*} \alpha = \sum_{k=0}^{m-1} a_k \zeta^k, \quad a_k \in K. \end{align*} Applying $\phi$: \begin{align*} \phi(\alpha) = \phi\!\left(\sum_{k=0}^{m-1} a_k \zeta^k\right) = \sum_{k=0}^{m-1} \phi(a_k) \cdot \phi(\zeta)^k = \sum_{k=0}^{m-1} a_k \cdot \zeta^k = \alpha. \end{align*} Here we used three properties of $\phi$: additivity ($\phi$ respects sums), multiplicativity ($\phi(\zeta^k) = \phi(\zeta)^k$), and the fact that $\phi$ fixes $K$ pointwise ($\phi(a_k) = a_k$ for $a_k \in K$, since $\phi \in \operatorname{Gal}(L/K)$). Combined with the hypothesis $\phi(\zeta) = \zeta$, every term is unchanged, so $\phi(\alpha) = \alpha$ for all $\alpha \in L$. This is the structural reason injectivity holds: the Galois group acts faithfully on $\zeta$ because $\zeta$ generates the entire extension. If $L$ were not generated by $\zeta$ alone (for instance, if we adjoined two independent roots), an automorphism could fix $\zeta$ while permuting the other generators nontrivially, and injectivity would fail. [/guided] [/step] [step:Deduce that $\operatorname{Gal}(L/K)$ is abelian and $[L:K] \mid \varphi(n)$] Since $\chi: \operatorname{Gal}(L/K) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times$ is an injective group homomorphism, $\operatorname{Gal}(L/K)$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$. The group $(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian (since multiplication in $\mathbb{Z}/n\mathbb{Z}$ is commutative). Every subgroup of an abelian group is abelian, so $\operatorname{Gal}(L/K)$ is abelian. For the divisibility: $L/K$ is a finite extension (since $\zeta$ is algebraic over $K$, being a root of $t^n - 1 \in K[t]$), so $|\operatorname{Gal}(L/K)| \le [L : K] < \infty$. The injectivity of $\chi$ gives $|\operatorname{Gal}(L/K)| \le |(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n)$. Since $L/K$ is a splitting field of the separable polynomial $t^n - 1$ over $K$ (separability holds because $\operatorname{Char} K \nmid n$ ensures $\gcd(t^n - 1, nt^{n-1}) = 1$ in $K[t]$, so $t^n - 1$ has no repeated roots), the extension $L/K$ is Galois, and therefore $|\operatorname{Gal}(L/K)| = [L : K]$. By Lagrange's theorem applied to the injection $\chi$, the order of the subgroup $\operatorname{im}(\chi)$ divides $|(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n)$. Since $|\operatorname{im}(\chi)| = |\operatorname{Gal}(L/K)| = [L : K]$ (using injectivity and the Galois property), we conclude $[L : K] \mid \varphi(n)$. [guided] This step draws the two stated consequences from the established injection. Each relies on a different property of injective group homomorphisms. **Abelianness.** An injective homomorphism $\chi: G \hookrightarrow A$ into an abelian group $A$ embeds $G$ as a subgroup of $A$. Since $(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian — the multiplication $\bar{i} \cdot \bar{j} = \overline{ij} = \overline{ji} = \bar{j} \cdot \bar{i}$ is commutative — every subgroup of $(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian. In particular, $\operatorname{Gal}(L/K) \cong \operatorname{im}(\chi) \le (\mathbb{Z}/n\mathbb{Z})^\times$ is abelian. This abelianness is notable because the Galois group of a general finite extension need not be abelian (e.g., $\operatorname{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong S_3$ for a root $\alpha$ of an irreducible cubic with discriminant that is not a perfect square). The special structure of cyclotomic extensions — generated by roots of unity, with the Galois action determined by exponentiation — forces commutativity. **Divisibility.** We must show $[L : K] \mid \varphi(n)$. This requires establishing that $L/K$ is Galois (so that $|\operatorname{Gal}(L/K)| = [L:K]$), and then applying Lagrange's theorem. The extension $L = K(\zeta)$ is the splitting field of $t^n - 1$ over $K$: all $n$-th roots of unity are powers of $\zeta$, so $t^n - 1$ splits completely in $L[t]$. We verify separability: the formal derivative of $t^n - 1$ is $nt^{n-1}$. The hypothesis $\operatorname{Char} K \nmid n$ ensures $n \neq 0$ in $K$, so $\gcd(t^n - 1, nt^{n-1}) = \gcd(t^n - 1, t^{n-1})$. Since $t^{n-1} \mid t^n$ but $t^{n-1} \nmid -1$ (as $\deg(-1) = 0 < n-1$ for $n \ge 2$; the case $n = 1$ is vacuous since $L = K$), the gcd is $1$. Therefore $t^n - 1$ has no repeated roots in any extension of $K$, confirming separability. Since $L/K$ is a splitting field of a separable polynomial, $L/K$ is Galois, and $|\operatorname{Gal}(L/K)| = [L : K]$. Now, the injection $\chi$ gives $\operatorname{im}(\chi) \le (\mathbb{Z}/n\mathbb{Z})^\times$. By Lagrange's theorem, the order of any subgroup divides the order of the ambient group: \begin{align*} |\operatorname{im}(\chi)| \mid |(\mathbb{Z}/n\mathbb{Z})^\times| = \varphi(n). \end{align*} Since $\chi$ is injective, $|\operatorname{im}(\chi)| = |\operatorname{Gal}(L/K)| = [L:K]$. Therefore $[L:K] \mid \varphi(n)$. [/guided] [/step]