Let $f\in\mathbb{Q}[x]$ and let $E$ be its seperable field over $\mathbb{Q}$. Suppose that $\text{Gal}(E/F)=D_{2017}$. Show that $f$ is solvalbe.
I would like to verify my answer, please.
We know that $f$ is solvable if and only if $G_f=\text{Gal}(E/F)$ is solvable. Consider $H=C_{2017}$. $H\triangleleft G$ because for all $g=a^ib^j\in D_{2017}$ where $0\leq 0<2017, j=0,1$, and for all $h=a^k\in H$ where $0\leq k<2017$, we have that $g^{-1}hg\in H$. According to the cardinals of $G$ and $H$ we have $G\diagup H\cong C_2$ which is commutative.
Also, $C_{2017}\diagup \{1\}\cong C_{2017}$ which is commutative, too. Thus $G_f$ is solvable and $f$ is solvable too.
Since every dihedral group $D_n=\langle r,s \rangle$ for $n\ge 3$ is $2$-step solvable, $Gal(E/F)$ is solvable and hence $f$ is solvable. In fact $$ D_n'=\langle r^2\rangle, \; D_n''=\{id\}. $$ Another way to show that $D_n$ is solvable, is that $C_n$ is a normal abelian subgroup of $D_n$ with abelian quotient $D_n/C_n\cong C_2$. Since both $C_2,C_n$ are solvable, so is $D_n$.