We know that for a Galois extension $E/F$ its Galois group $\text{Gal}(E/F)\simeq \varprojlim(F_i/F)$ where $\left \lbrace F_i \right \rbrace_{i\in I}$ is the family of all finite Galois extension of $F$. Therefore let $q$ be a power of a prime. For a finite field $\mathbb{F}_q$ its Galois group of its algebraic closure $\text{Gal}(\bar{\mathbb{F}}_q/\mathbb{F}_q\simeq\varprojlim\mathbb{Z}/n\mathbb{Z})=\varprojlim\mathbb{Z}/n!\mathbb{Z})$(for we can insert $\mathbb{F}_{q^n}$ into a sufficiently large field $\mathbb{F}_{q^{m!}}$).
And my question is that what is the Galois group of $\bar{\mathbb{Q}}/\mathbb{Q}$? Because all finite groups can be inserted into permutation group $S_n$, so I guess that $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\simeq S_{\infty} $ where $S_{\infty}:=\left \lbrace f:\mathbb{N}\rightarrow \mathbb{N}:f \text{ is a bijective} \right \rbrace$. I think this idea may be incorrect, could someone tell me the right answer?