Inverse Galois Problem

$\mathbf{Problem:}$ Use the fact that any finite group $G$ is isomorphic to a subgroup of $S_n$ to prove that given any finite group $G$ there exists fields $F$ and $E/F$ such that

$\displaystyle \text{Gal }E/F\cong G$

$\mathbf{Proof:}$ Any field $K$, $x_1,x_2,\cdots,x_n$ are indeterminate. Let $E=K(x_1,x_2,\cdots,x_n)$. $D$ is the field of symmetric rational functions in $latex x_1,x_2,\cdots,x_n$ Or if we let $p_1,p_2,\cdots,p_n$ are basic symmetric polynomials, namely

$\displaystyle p_1=\sum x_i$, $\displaystyle p_2=\sum _{i>j} x_ix_j$, $\cdots \displaystyle p_n=x_1x_2\cdots x_n$,

then $D=K(p_1,p_2,\cdots,p_n)$.

As we all know, $\text{Gal }E/D=S_n$. Since $G\leq S_n$, by the Galois corresponding theorem, there exists an intermediate field $F$ such that $\text{Gal }E/F\cong G$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ This is the easiest inverse Galois problem since one does not specify the ground field. If you restrict $F=\mathbb{Q}$, things are much trickier.