I wish to prove that if $F / \mathbb Q$ is a finite Galois extension such that $|Gal(F / \mathbb Q)|$ is odd, then $F \subseteq \mathbb R$. I am skeptical of my current proof since my peers have told me that they invoked some Sylow Theory, whereas my proof (below) didn't. I was wondering if someone can provide me some correctness.
$\textbf{EDIT:}$ I have updated the proof below as of July 12, 2023 implementing some of the corrections pointed out. Feel free to comment.
$\textbf{Proof: }$
Let $F\subseteq \mathbb C$ such that $F / \mathbb Q$ is a finite Galois extension. Suppose that $\left\lvert Gal(F / \mathbb Q) \right\rvert$ is odd. By the definition of a Galois
extension, since $F / \mathbb Q$ is Galois, $F / \mathbb Q$ is separable.
Hence by the Primitive Element theorem, $F = \mathbb Q(z)$ for some $z \in F$, so that $\mathbb Q(z) / \mathbb Q$ is a finite Galois extension. By the Characterization Theorem we have that $\left\lvert Gal(F / \mathbb Q) \right\rvert = [F : \mathbb Q]$.
It follows that
\begin{equation}
[F : \mathbb Q] = [\mathbb Q(z) : \mathbb Q] \tag{1}
\end{equation}
and hence $[\mathbb Q(z) : \mathbb Q]$ is odd. Let $m_{z}(x) \in \mathbb Q[x]$ be the minimal polynomial of $z$ over $\mathbb Q$. By (1), since $[\mathbb Q(z) : \mathbb Q] = \deg m_z(x)$ we have
that $\deg m_{z}(x)$ is odd. Note that since $\mathbb Q(z) / \mathbb Q$ is Galois it is normal, and by the Normality Theorem, $m_{z}(x)$ splits over $\mathbb Q(z)$ since $m_z(x)$ has a root in $\mathbb Q(z)$ (namely $z$). Moreover, by minimality it must be the case that $\mathbb Q(z)$ is the splitting
field of $m_z(x)$ over $\mathbb Q$ (see Appendix for a proof). We claim that $m_{z}(x) \in \mathbb Q[x]$ has a real root in $\mathbb Q(z)$. Indeed, all polynomials $f(x) \in \mathbb R[x]$ of odd degree have a real
root (see Appendix for a proof). Let $q \in \mathbb R$ be a real root of $m_z(x)$. Then $\mathbb Q(z) = \mathbb Q(q) \subseteq \mathbb R$.
$\Large{Appendix}$
$\textbf{Claim 1:}$
The minimal polynomial $m_{z}(x) \in \mathbb Q[x]$ for $z$ over $\mathbb Q$ has a real root.
$\textit{Proof of claim:}$
Let $n = \deg m_z(x)$. Then $n$ is odd as we have established in the previous paragraph. Moreover, since $m_{z}(x)$ is irreducible and
$\mathbb Char(\mathbb Q) = 0$, we have that $m_z$ is separable, and hence has distinct roots. Suppose for a contradiction that the roots of $m_z(x)$ are purely
complex. Let $X \subseteq\mathbb C \setminus \mathbb R$ be the set of roots of $m_z(x)$ where $n = \left\lvert X \right\rvert$.
Let $\alpha \in X$ be a root of $m_z(x)$. Then
\begin{equation}
m_z(\alpha) = 0 \text{ and } m_z(\overline{\alpha}) = 0
.\end{equation}
But then this would imply that $n = \left\lvert X \right\rvert$ is even, a
$\textbf{Claim 2:}$
The field $\mathbb Q(z)$ is the splitting for $m_z(x)$ over $\mathbb
$\textit{Proof of claim:}$
We have already established that $m_z(x)$ splits over $\mathbb Q(z)$ since $m_z(x)$ has a root in $\mathbb Q(z)$ (namely $z$). Let $\{\alpha_1,\dots,\alpha_n\}$ be the full set
of roots of $m_z(x)$ where $\alpha_1 = z$. Then $K = \mathbb Q(\alpha_1,\dots,\alpha_n)$ is the splitting field of $m_z(x)$ over $\mathbb Q$.
Since $m_z(x)$ splits over $\mathbb Q(z)$, it must be the case that $\alpha_1,\dots,\alpha_n \in \mathbb Q(z)$. By minimality $K \subseteq \mathbb Q$. Since $z = \alpha_1 \in K$ by
minimality $\mathbb Q(z) \subseteq K$. Hence $K = \mathbb Q(z)$.
Best Answer
If $F$ is Galois over $\Bbb Q$, then $\vert \operatorname{Aut}(F/ \Bbb Q) \vert = [F: \Bbb Q]$. If the minimal polynomial of $z$ has a non-real root, then $\sigma: F \to F \in \operatorname{Aut}(F/ \Bbb Q)$, where $\sigma(\alpha) = \bar{\alpha}$ is complex conjugation. But $\vert \sigma \vert = 2$, which would mean $2 \mid \vert \operatorname{Aut}(F/ \Bbb Q) \vert=[F:\Bbb Q]$, contrary to our hypothesis that $F$ has odd degree over $\Bbb Q$.