Absolute Galois group

Remarks

It does not depend on the choice of $K^{s}$ up to isomorphism, because two separable closures are isomorphic over $K$. The induced isomorphism of Galois groups is unique only up to an inner automorphism. If $K$ is perfect, for example of characteristic $0$, then $K^{s}$ is an algebraic closure of $K$. Example: $G_\R = \Gal(\C/\R)$ has order $2$, generated by complex conjugation, since $\C$ is algebraically closed and $[\C : \R] = 2$. Example: for a finite field $\F_q$, the group $G_{\F_q}$ is isomorphic to the profinite completion $\hat{\Z}$ of $\Z$, topologically generated by $x \mapsto x^q$. Non-example: $\Gal(\Q(i)/\Q)$ is not the absolute Galois group of $\Q$, since $\Q(i)$ is not separably closed; $x^2 - 2$ has no root in it. If $K$ is separably closed, then $G_K$ is trivial.