Gauss-Wantzel theorem asserts that a polygon with $n$ sides is constructible if and only if $n$ is a product of a power of $2$ and distinct prime Fermat numbers, where the Fermat number of index $k$ is $F_{k}=2^{2^k}+1$. Conjecturally, the only prime Fermat numbers are those with index $k\leq 4$.

Kinda coincidentally, those values of $k$ are also those for which the symmetric group of index $k$ and order $k!$ is solvable. My question is thus: is it really a coincidence? Or can we exhibit a rather natural structural similarity between a sequence of constructible polygons the order (as number of sides) of each dividing the order of the next one and a sequence of subgroups of $S_4$ to show that there is no prime Fermat number of index at least $5$? In other words, can we reasonably expect to prove in a Galois theoretic framework that $F_{k}$ for $k$ positive is prime if and only if the general polynomial equation of degree $k$ is solvable by radicals?

Thanks in advance.

Edit: As $S_{4}$ is solvable, there exists a normal series with abelian factor groups $\{e\}=G_{0}<...<G_{4}=S_{4}$. The idea is to map $G_{i}$ to a product $N_{i}$ of $n+1$ Fermat primes such that $N_{i}\mid N_{i+1}$.