$\zeta_{15}$ is $15$th primitive $n$th root of unity.
Question: Find the structure of the group $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$
I know that if $p$ is prime then $G=Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.
Does my required group have any relation to some cyclic group $\mathbb{Z_n}$, i.e. abelian
Would really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks
Best Answer
Let $G$ be the Galois group of $\Bbb Q(\zeta_n)$ over $\Bbb Q$. An element $f \in G$ is entirely determined by its image on $\zeta_n$.
Since $f(\zeta_n)^k=f(\zeta_n^k)=1 \iff \zeta_n^k=1$ (recall that $f$ is a field automorphism), you know that $f(\zeta_n)$ is a primitive $n$-th root of unity: $$f(\zeta_n)=\zeta_n^{k_f}$$ for some integer $1≤k_f≤n$ coprime with $n$.
Therefore, you have a well-defined map $$\alpha : G \to (\Bbb Z/n\Bbb Z)^* \qquad \alpha(f)=[k_f]_n$$ You can check that this is a group isomorphism. It is clearly injective. Since $|G|=[\Bbb Q(\zeta_n):\Bbb Q]=\text{deg}(\Phi_n)=\phi(n) = |(\Bbb Z/n\Bbb Z)^*|$, $\alpha$ is bijective.
Finally, $(f \circ g)(\zeta_n)=f(g(\zeta_n)) = f(\zeta_n^{k_g})=\zeta_n^{k_g \, k_f}$ shows that $$\alpha(f \circ g) = [k_{f \circ g}]_n = [k_f]_n[k_g]_n=\alpha(f)\alpha(g).$$
More generally, let $K$ be a field of characteristic coprime with $n$ (e.g. if $\text{car}(K)=0$), and suppose that $R_n = \{x \in \overline K \mid x^n=1\}$ denotes the set of $n$-th roots of unity in an algebraic closure $\overline K$ of $K$ (notice that $R_n$ is always a cyclic group for the multiplication, since it is a finite subgroup of $(K^*,\cdot)$).
Then the extension $K(R_n)$ over $K$ is Galois (it is separable because $n$ is coprime with the characteristic of $K$) and you can show similarly that the Galois group of $K(R_n)$ over $K$ embeds in $(\Bbb Z/n\Bbb Z)^*$.