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Been reviewing some Galois theory, and computing Galois groups has been relatively routine, but one has me stumped.

Suppose $\omega$ is a primitive 37th root of unity. Let's set $\alpha=\omega+\omega^{10}+\omega^{26}$. I want to compute the Galois group of $\mathbf{Q}(\alpha)/\mathbf{Q}$.

I realize that $\mathbf{Q}(\alpha)\subseteq\mathbf{Q}(\omega)$, so $[\mathbf{Q}(\alpha):\mathbf{Q}]$ must divide $[\mathbf{Q}(\omega):\mathbf{Q}]=\varphi(37)=36$. So the order of the Galois group must be a divisor of 36, but that's about all I've managed.

What's the trick to finding this Galois group? Thanks!

Edit: Any automorphism has to map $\omega$ to $\omega^k$ where $(k,37)=1$. I noticed that the maps sending $\omega\mapsto\omega^{10}$ and $\omega\mapsto\omega^{26}$ both fix $\alpha$. Where does one go from there?

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    @lhf Finally got it, thanks very much for your kind help.2011-12-01

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You have pretty much answered your own question! Indeed, suppose $\sigma \in \text{Gal}(\mathbf{Q}(\omega)/\mathbf{Q})$ is such that $\sigma \alpha = \alpha$. As you noted, $\sigma \omega = \omega^k$ for some $k$ with $(k, 37)=1$. Then we have $\omega^k + \omega^{10k} + \omega^{26k} = \omega+\omega^{10}+\omega^{26}$. This implies $\{k,10k,26k\}=\{1,10,26\}$ mod $37$, by the linear independence of $\{1, \omega, \dots, \omega^{36}\}$. In particular, we have either $k=1, 10k=1$ or $26k=1$. Hence $k=1, k=10$ or $k=26$, and a quick check shows that all of these work. Therefore...

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    @MathMastersStudent Because the group is $(\mathbb{Z}/37\mathbb{Z})^\times$, which is abelian (in fact cyclic of order 36). This is a basic result in the theory of cyclotomic fields. :)2011-12-06