Galois extension and prime number. Let $G$ be a finite group with order $n$, i.e., $|G|=n$. Show that there is a prime number $p\geq n$ and a finite Galois extension $L/K$ with $Gal(L/K)\approx G$ and $[K:\mathbb{Q}]=p!/n$.
Honestly, I have no idea!
 A: Consider $G$  as a subgroup of $S_n$, the symmetric group (have $G$ act on itself by left multiplication - $n$ is the order of $G$). The group $S_n$ sits inside of $S_p$, for any prime $p$ larger  than  or equal to $n$. Now, the inverse Galois problem over $\mathbb Q$ is solved for the symmetric group ( according to https://en.wikipedia.org/wiki/Inverse_Galois_problem, by Hilbert - but see below).  So there exists a Galois extension $L/ \mathbb  Q$ with group $S_p$.  Let $K= L^G$, the fixed field of $G$. Then $p! = [L:{\mathbb Q}] = [L:K] \ [ K:{\mathbb Q}] = n\  [ K:{\mathbb Q}]$. Hence $$G = \mathop{\rm Gal} (L/K )$$ and
$$[K:{\mathbb Q}] = p! /n. $$  


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*For a solution to the inverse Galois problem over $\mathbb Q$ for $S_p$ with $p$ prime $\ge 5$ (courtesy of Brauer), see section 4.10 of Basic Algebra I (surely one of the most discouraging text book titles ever) of Jacobson, Second Edition. The section also refers to a exercise 5, on page 305, which follows a different argument due Tate to solve the inverse Galois problem for $S_n$ over $\mathbb Q$.

