# Why is the Galois group for a cyclotomic extension isomorphic to $Z_n^{\times}$ and not to $Z_{n-1}$?

I'm struggling to understand this. Let $\mathbb{Q}(\xi)$ be the splitting field for $x^n - 1$, so here $\xi$ is the primitive n-th root of unity. If I consider the possible automorphisms of this field, I will get:

$$\iota \textrm{ (the identity)}, \qquad \textrm{and} \qquad \sigma_k: \xi \mapsto \xi^k, \quad2 \leq k< n.$$

I will have $n-1$ elements, same as $Z_{n-1}$. How can it be shown that this is actually $Z_n^{\times}$ instead? What am I missing?

• Why do you have $n-1$ elements? Can $\xi$ be mapped to any $\xi^k$ or just those $k$'s relatively prime to $n$? Commented May 10, 2016 at 15:29
• Write down the composition explicitly. For an example of what can go wrong, take $n = 4$. Commented May 10, 2016 at 15:51
• Every $\mathbb Q$-automorphism of $\mathbb Q(\zeta)$ is a permutation of the set $\{1,\zeta,\zeta^2,\zeta^3,\dots\}$ that fixes 1 and preserves the multiplicative structure of $\langle \zeta \rangle \cong \mathbb Z / n \mathbb Z$, as a multiplicative subgroup of $\mathbb C$. So $\mathrm{Gal}(\mathbb Q(\zeta) / \mathbb Q)$ is contained in $\mathrm{Aut}(\mathbb Z / n \mathbb Z) \cong (\mathbb Z / n \mathbb Z)^\times$. Order consideration will get you equality. Commented May 10, 2016 at 16:13

Simply, not all the maps $$\sigma_k : \xi \mapsto \xi^k$$ are field automorphisms. For example, consider $$n = 4$$; in this case, we usually denote the primitive root by $$i$$. Then, $$\sigma_2$$ maps both $$\pm i$$ to $$-1$$, so $$\sigma_k$$ is not an automorphism of $$\Bbb Q[i]$$ fixing $$\Bbb Q$$.
On the other hand, any automorphism must permute the roots of unity, so the map $$\xi \mapsto \xi^k$$ must be invertible if it is to be an automorphism, but this only holds if $$k, n$$ are coprime, that is, if $$k \in \Bbb Z_n^{\times}$$. Note that (for $$n > 1$$) $$\Bbb Z_n^{\times} \cong \Bbb Z_{n - 1}$$ iff $$k, n$$ are coprime for all $$1 \leq k < n$$, that is, iff $$n$$ is prime.
Again for $$n = 4$$ we have $$\Bbb Z_4^{\times} = \{1, 3\}$$, so the only two automorphisms are characterized by $$i \mapsto i$$ (the identity) and $$i \mapsto i^3 = -i$$ (complex conjugation).
• Another reason is that the degrees are wrong. The degree of the extension gotten by adjoining an $n$-th root of unity $\zeta$ is equal to the degree of the minimal polynomial for $\zeta$, and this degree is not $n-1$, but $\varphi(n)$, the Euler number of integers $\le n$ that are relatively prime to $n$. Commented Jul 6, 2021 at 1:32