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Let $K$ be a field with $\mbox{char}(K)=0$. I know that if $\xi$ is a primitive $n$-th root of the unity and $K(\xi)\big/K$ is a cyclotomic extension, then $\mbox{Gal}\left(K(\xi)\big/K\right)$ is isomorphic to a subgroup of $\mathbb Z_n^*$ (the multiplicative group).

However, I've found in some examples that they use that $\mbox{Gal}\left(K(\xi)\big/K\right)$ is isomorphic to $\mathbb Z_n^*$. I wonder when this is possible.

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I'm not sure what you meant with "the multiplicative group without the zero", but $\,\left(\Bbb Z/n\Bbb Z\right)^*\,$ is not the original group (ring, in fact) without the zero, but the multiplicative group of all units modulo $\,n\,$. Unless $\,n\,$ is a prime, this is less than the non-zero elements. – DonAntonio Jan 26 '13 at 15:53
+1 Well, that helped indeed. Could you provide an example, please? – synack Jan 26 '13 at 16:04
An example is furnished by any $n$ that’s not a prime. Like $n=4$. – Lubin Jan 26 '13 at 17:25
So, it would be $\mathbb (Z/4\mathbb Z)^* = \left\{1,\xi, \xi^2\right\}$, right? – synack Jan 26 '13 at 17:32
No, it is $\,\left(\Bbb Z/4\Bbb Z\right)^*=\{1,3\}\,$ , multiplicatively. A group of order two. – DonAntonio Jan 26 '13 at 17:56
up vote 3 down vote accepted

$\mathrm{Gal}(K(\xi)/K)\cong (\mathbb{Z}/n\mathbb{Z})^*$ is true when $K=\mathbb{Q}$. More generally, it is true if the $n$-th cyclotomic polynomial is irreducible over $K$.

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For $n$ a prime, and $K = \mathbf{Q}$, you will indeed get $\mathrm{Gal}(\mathbf{Q}(\xi)/\mathbf{Q}) \cong \mathbf{Z}_n \setminus \{ 0 \}$.

But then if you understand $\mathbf{Z}_n^{\star}$ to be the group of invertible elements of (the multiplicative monoid of) $\mathbf{Z}_n$, then $\mathrm{Gal}(\mathbf{Q}(\xi)/\mathbf{Q}) \cong \mathbf{Z}_n^{\star}$ for all $n$.

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