Every finite group is isomorphic to some Galois group for some finite normal extension of some field.
Solution 1:
Your idea is right, and in fact it is the standard proof: let $G$ be a given group, and let $n=|G|$.
There is a field extension $K/E$ with $\mathrm{Aut}_E(K)\cong S_n$ (e.g., $E=\mathbb{Q}(s_1,\ldots,s_n)$, $K=\mathbb{Q}(x_1,\ldots,x_n)$, where $s_1,\ldots,s_n$ are the symmetric polynomials in $x_1,\ldots x_n$).
By Cayley's Theorem, we know that $G$ is isomorphic to a subgroup $H$ of $S_n$, so if we let $L$ be the fixed field of $H$ in $K$, then by the Fundamental Theorem of Galois Theory we know that $K/L$ is Galois and $\mathrm{Aut}_{L}(K) = H\cong G$.