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The problem I have is the following:

Show that $\mathbb Q(\zeta_{49},\sqrt{7})=K $ is a Galois extension and determine the Galois' group.

I know that is a Galois extension because it is the splitting field of $f(x)=(x^2-7)(\phi_{49}(x))$ , where $\phi_{49}(x)$ is the minimal polynomial of $\zeta_{49}$ (a $49$th primitive root of unity) but I have difficulty deciding if $\sqrt{7} \in \mathbb Q(\zeta_{49})$.
I know that $Aut(\mathbb Q(\zeta_{49})/\mathbb Q)\cong \mathbb Z_{49}^*\cong \mathbb Z_{42}$ is a cyclic group, so there is an unique subextension of $K$ that has grade 2 on $\mathbb Q$. But I have no idea how to conclude.

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    $\begingroup$ Notice that only $7$ is ramified in $\Bbb Q(\zeta_{49})$. $\endgroup$ – Watson Jan 10 '17 at 19:51
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    $\begingroup$ You can have a look to this. $\endgroup$ – Watson Jan 10 '17 at 20:02
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    $\begingroup$ Theorem 3.41, p 62 in these course notes might be of interest: J.S. Milne, Algebraic Number Theory. Version 3.06. May 28, 2014. jmilne.org/math/CourseNotes/ANT.pdf $\endgroup$ – punctured dusk Jan 10 '17 at 20:12
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    $\begingroup$ A non-trivial way to solve your problem is the following: if $\sqrt 7 \in L:=\Bbb Q(\zeta_{49})$, then $K:=\Bbb Q(\sqrt 7) \subset L$. But $K$ has discriminant $28$, and $2$ is what we call a ramified prime in $K$. Thus it would also be ramified in $L$, but this is not the case. The only ramified prime in $\Bbb Q(\zeta_{p^n})$ is $p$ (for $p>2$ prime). Conclusion : $\sqrt 7 \not \in L=\Bbb Q(\zeta_{49})$. $\endgroup$ – Watson Jan 10 '17 at 20:27
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    $\begingroup$ WimC has shown that $\mathbb{Q}(\sqrt{7}) \cap \mathbb{Q}(\zeta_{49}) = \mathbb{Q}$, so this result finishes the computation of the Galois group. $\endgroup$ – André 3000 Jan 10 '17 at 20:51
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The degree $[\mathbb{Q}(\zeta_{49})\cap \mathbb{R}:\mathbb{Q}]=21$ is odd, so...

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    $\begingroup$ You mean \cap \Bbb R, I suppose. $\endgroup$ – Watson Jan 10 '17 at 20:31
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    $\begingroup$ Note that $\Bbb{Q}[\zeta_{49}] \cap \Bbb{Q} = \Bbb{Q}$. $\endgroup$ – Vik78 Jan 10 '17 at 20:31
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    $\begingroup$ @Watson I would agree. :-) $\endgroup$ – WimC Jan 10 '17 at 20:31
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    $\begingroup$ @MarcoLecci, In general, $[\mathbb{Q}(\zeta_n)\cap \mathbb{R}:\mathbb{Q}]=\varphi(n)/2$. You can show that $\mathbb{Q}(\zeta_n)\cap \mathbb{R}$ is generated by $\zeta_n+\zeta_n^{-1}$ and that $\zeta_n$ has degree $2$ over $\mathbb Q[\zeta_n+\zeta_n^{-1}]$. (Or use the more elegant argument given by SpamIAm; probably you need to use it anyway :) ) $\endgroup$ – punctured dusk Jan 10 '17 at 20:38
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    $\begingroup$ @MarcoLecci The maximal real subfield $\mathbb{Q}(\zeta_{49}+\zeta_{49}^{-1})$ is the fixed field of complex conjugation, hence $[\mathbb{Q}(\zeta_{49}):\mathbb{Q}(\zeta_{49}+\zeta_{49}^{-1})] = 2$. $\endgroup$ – André 3000 Jan 10 '17 at 20:40
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Another approach than @WimC’s good answer: $[\Bbb Q(\zeta_{49}):\Bbb Q]=42$, and it’s an extension ramified only at $7$, with cyclic Galois group. So the extension has only one intermediate field of each possible degree, every one of them ramified only at $7$. The quadratic extension ramified only at $7$ is $\Bbb Q(\sqrt{-7}\,)$.

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The quadratic Gauss sum $$S=\sum_{a\in\mathbb F_p}\zeta_p^{a^2}$$ for odd prime $p$ satisfies $$S^2=(-1)^{(p-1)/2}p$$ as shown by direct calculation. It follows for $p=7$ that $$S=\pm\,i\sqrt7\in\mathbb Q(\zeta_7)\subset\mathbb Q(\zeta_{49}),$$ so $\mathbb Q(i\sqrt7)$ and not $\mathbb Q(\sqrt7)$ is the unique quadratic subfield of $\mathbb Q(\zeta_{49})$.

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