Let $G$ be a group with a normal subgroup $N$ of order $7$ such that $G/N$ is isomorphic to the dihedral group of order $10$. Prove the following:
(a) $G$ has a normal subgroup of order $35$.
(b) $G$ has $5$ non-normal subgroups of order $14$.
I think I have some ideas for this problem, but I'm not sure if they are complete/formal enough. Here are my thoughts:
(a) Since $G/N \cong D_{10}$, the order of the quotient group is $10$. Since $N$ has order $7$ in $G$, we may conclude by Lagrange's theorem that $|G| = 70 = 2\cdot5\cdot7$. By an immediate application of the theorems of Sylow, $G$ has a single Sylow $7$-subgroup, $P_7$. This subgroup is normal. Furthermore, there exists some Sylow $5$-subgroup of $G$ which is not necessarily normal we can denote as $P_5$. Hence the group $P_7P_5$ has order $\frac{|P_7||P_5|}{|P_7 \cap P_6|}$, but the intersection of Sylow $p$-subgroups for distinct $p$ is trivial, so the product of these two groups is a group of order $35$. Since it has order $35$, it has index $2$ in $G$, so it is normal in $G$.
(b) This is the one I am less confident about. In $D_{10}$, the subgroups generated by the $5$ reflection symmetries across the vertices of the polygon are all non-normal, and have index $5$. By the correspondence theorem, there are subgroups in $G$ corresponding to these subgroups, with index $5$, meaning they have order $14$.
I don't think I have said this formally, and may be missing some details, so I would appreciate any comments or solutions.