well the question is in the title. I asked myself this question while thinking about something in grothendieckteichmüller theory. I guess class field theory gives some insight into this, or i am missing something absolutely obvious..

The proof of triviality is a step in the famous NeukirchUchida theorem of anabelian geometry, which says a number field is characterized by its absolute Galois group, even functorially, in an appropriate sense. The key elementary fact is the following: Let $k$ be a number field, $K$ an algebraic closure, and $G=Gal(K/k)$. Let $P_1$ and $P_2$ be two distinct primes of $K$ with corresponding decomposition subgroups $G(P_i)\subset G$. Then $G(P_1)\cap G(P_2)=1.$ Once this is stated for you, it's essentially an exercise to prove. Determining the center of $G$ becomes then completely straightforward: Suppose $g$ commutes with everything. Then for any prime $P$, $G(gP)=gG(P)g^{1}=G(P)$. So $g$ must fix every prime, implying that it's trivial. I think this is spelled out in the book Cohomology of Number Fields, by Neukirch, Schmidt, and Wingberg. Unfortunately, I left my copy on the plane last year. 

