Let $p$ be a prime not equal to $2$. Let $f(x)$ be an irreducible polynomial over $\mathbb{Q}$ of degree $p$ with Galois group isomorphic to the dihedral group $D_{2p}$. I need to show that $f(x)$ has all real roots or exactly one real root. (Note that $D_{2p} = <s,r : s^2 = r^p = 1, rs = sr^{-1}>$).
I'm very much stumped and don't even know where to begin. Can I get any aid or hints?
On
Take $g\in Gal(f) \cong D_{2p}$ of order $p$, it must permute the $p$ roots $x_0,\ldots,x_{p-1}$ of $f$ transitively (since otherwise it would be of order $<p$) so wlog $g(x_j) = x_{j+1}$ (with $j$ taken $\bmod p$). The dihedral group is generated by one more element $h$ of order $2$ and which satisfies $hgh = g^{-1}$.
Let $h(x_j) = x_{e_j}$, then $g^{-1}(x_{j+1}) =x_j =h(x_{e_j})= hgh(x_{j+1})=hg(x_{e_{j+1}})=h(x_{e_{j+1}+1})$ so $x_{e_{j+1}+1} = x_{e_j}$ and $e_{j+1} = e_j-1$ ie. $e_j = e_0-j, h(x_j) = x_{e_0-j}$.
If the roots are not all real we can take $h$ to be the complex conjugaison so that $x_j$ is real iff $h(x_j) = x_j$ iff $2j = e_0 \bmod p$.
If $f$ has at one non-real root, then one $\alpha\in G$ is given by complex conjugation. If $x_1,x_2$ are two real roots, then they are fix under $\alpha$. By transitivity of $G$, there exists $\beta\in G$ with $\beta(x_1)=x_2$. Then $\beta^{-1}\alpha\beta$ is of order two and leaves $x_1$ fix. The product of two distinct reflections in $D_{2p}$ is a rotation (i.e., of order $p$), hence $\alpha\beta^{-1}\alpha\beta$ is of order $p$ and has a fixpoint. Hence it can permute only the other $p-1$ roots. But $S_{p-1}$ has no element of order $p$.