# Orders of field automorphisms of algebraic complex numbers

Let $Aut(\bar{Q})$ be the automorphism group on the field of algebraic complex numbers. The order of an element $f \in Aut(\bar{Q})$ is the least natural number $n$ (if there exists one) such that $(f)^{n}$ is identity. What are the possible orders of elements of $Aut(\bar{Q})$?

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if $K/F$ is a field extension with $K$ algebraically closed and $[K:F] < \infty$, then $[K:F] = 1$ or $2$, and in the latter case, $F$ is real-closed.
Thus the answer here is that $n$ can be $1$, $2$ or $\infty$, and all possibilities occur: the field of real algebraic numbers gives an index $2$ subfield of $\overline{\mathbb{Q}}$.
(Also, just to be sure, there are elements of infinite order! E.g., if not then every element would have order $1$ or $2$, so the absolute Galois group would be abelian, and thus every finite Galois group over $\mathbb{Q}$ would be abelian, and this is certainly not the case.)