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This question already has an answer here:

Let $\zeta_{21}$ be a primitive 21st root of 1. Describe explicitly the elements of the Galois group $\text{Gal}(\mathbb Q(\zeta_{21}):\mathbb Q)$ and determine the structure of this group.

From finding the minimal polynomial I know that the order of the Galois group must be 12. So $\text{Gal}(\mathbb Q(\zeta_{21}):\mathbb Q)≅(\mathbb Z/21\mathbb Z)^{\times}$.

I have seen online that $(\mathbb Z/21\mathbb Z)^{\times}$ has the structure $C_2 \times C_6$, how has this been found?

How can I describe the elements of this Galois group?

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marked as duplicate by Dietrich Burde, Rebellos, Misha Lavrov, user99914, user223391 Dec 2 '17 at 0:17

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  • $\begingroup$ How would you know just from the minimal polynomial that the galois group has order 12? $\endgroup$ – Gregory Grant Nov 30 '17 at 20:32
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I suppose you mean $U(21)$. This has $\phi(21)=12$ elements, and is not cyclic:

For what $n$ is $U_n$ cyclic?

From $\mathbb{Z}/21\cong \mathbb{Z}/3\times \mathbb{Z}/7$ we see that $U(21)\cong U(3)\times U(7)\cong C_2\times C_6$. Here we have used that $U(p)$ is cyclic of order $p-1$, see the above link.

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  • $\begingroup$ Okay I think I follow. And how can I describe the elements of this Galois group? I looked at the possible duplicate you mentioned but I found it a little confusing. $\endgroup$ – user484410 Nov 30 '17 at 21:03
  • $\begingroup$ Look up the proof for $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong (\mathbb{Z}/n)^{\times}$ (another duplicate here on MSE), which gives you a bijection between the automorphisms and units in $\mathbb{Z}/n$. This gives the description. $\endgroup$ – Dietrich Burde Nov 30 '17 at 21:05