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Let $M$ be a subfield of Complex field such that $M/\Bbb Q$ is a finite Galois extension. Show that if $[M:\Bbb Q]$ is an odd number, then $M$ is a subfield of Real field.

My current thought is since $M/\Bbb Q$ is Galois, it is a normal extension. That is, if $f(x)$ has a root in $M$, all roots of $f(x)$ are in $M$. However, I don't know how to continue from here.

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Since $M/\mathbb{Q}$ is finite and Galois, it can be thought of as the splitting field of some separable polynomial $p(x)$ with rational coefficients. If $M$ were not a subfield of $\mathbb{R}$, then $p(x)$ would have a nonreal complex root, $\alpha$. Since all of $p$'s coefficients are real then $\overline{\alpha}$ would be a root of $p$ as well. Now complex conjugation would be a nontrivial automorphism of $M$ fixing $\mathbb{Q}$, so it would be in the Galois group. But complex conjugation has order 2 as an automorphism, so the size of the Galois group, $[M:\mathbb{Q}]$ would be even by Lagrange's theorem.

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