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I've just started to study basics of Galois theory and I have some troubles with a simple task: I'm trying to prove that Galois group $Aut(\mathbb{Q}(\xi_n))$ is a subgroup of $(Z_n^{*}, \cdot)$

I know that roots of polynomial $x^n - 1$ go to conjugate elements under automorphism. I also know that we need to check that given set is closed under $(Z_n^{*}, \cdot)$ group operation and taking reverse element. But I can not even come up with an operation on $Aut(\mathbb{Q}(\xi_n))$, how can we define multiplication here? I'd be very grateful if you could clarify this bacics for me

grouplover
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    $Aut(\mathbb{Q}(\zeta_{n}))$ is isomorphic to $\mathbb{Z}{n}^{\times}$, not the subgroup of $\mathbb{Z}{n}^{\times}$. The operation of $Aut(\mathbb{Q}(\zeta_{n}))$ is the composition of functions – Nullstellensatz Aug 13 '24 at 13:09
  • Take $\sigma$ in $G=Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, $\sigma(\zeta_n)$ is an $n$-th primitive root of unity, so it can be written $\sigma(\zeta_n)=\zeta_n^k$ for some $k$ coprime to $n$. So you can introduce $\chi :\sigma \mapsto k$ from $G$ to $(\mathbb{Z}/n\mathbb{Z})^{\times}$, which is a well defined and injective group homomorphism. – Dietrich Burde Aug 13 '24 at 13:15
  • For example, we can define $\sigma_{i}(\zeta_{n})=\zeta_{n}^{i}$ and $\sigma_{j}(\zeta_{n})=\zeta_{n}^{j}$, where $(i,n)=(j,n)=1$. Then the composition of $\sigma_{i}$ and $\sigma_{j}$ sends $\zeta_{n}$ to $\zeta_{n}^{ij}$, which is similar to the operation $i\times j = ij$ in $Z_{n}^{\times}$. – Nullstellensatz Aug 13 '24 at 13:15

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