Galois group of algebraic closure of a finite field is Abelian The question is this:

Prove The Galois group of algebraic closure of a finite field is Abelian and the order of every element in the Galois group is infinite.

I have a trouble with the first part. How can I prove the Galois group is Abelian?
(Sorry, I'm very beginner in Galois theory.)
 A: Although I think the comments do nicely to answer your question, since you’ve asked for an expansion of my comment, here goes:
Let $k\supset \Bbb F_q$ be an extension of degree $n$, so that $\vert k\vert=q^n$, and every element of $k$ is a root of $f(X)=X^{q^n}-X$. This happens because the (cyclic) group $k^*$ has order $q^n-1$. Thus $k$ is all the roots of $f$, and only these.
Now $\varphi_q$, defined by $\varphi_q(z)=z^q$ is an $\Bbb F_q$-automorphism of $k$, in other words, an element of $\text{Gal}^k_{\Bbb F_q}$. You see, I think easily, that the first power of $\varphi_q$ to be identity on $k$ is the $n$-th power. So the powers of $\varphi_q$ fill out $\text{Gal}^k_{\Bbb F_q}$, in other words, this Galois group is cyclic, generated by $\varphi_q$.
Well, this is quite wonderful. The same formula, $\varphi_q(z)=z^q$, works no matter the size of $k$, i.e. no matter the value of $n=[k:\Bbb F_q]$. This formula even works on the algebraic closure $\Omega$ of $\Bbb F_q$, which is just the union of all the finite extensions of $\Bbb F_q$. I won’t go into detail here, but in some sense (fairly easily made precise, but I don’t want to go into details here), $\text{Gal}^\Omega_{\Bbb F_q}$ is topologically generated by $\varphi_q$, it is “pro-cyclic”. And definitely abelian.
( Maybe I might say that the significance of the very apposite comments of Lord Shark the Unknown and Jyrki Lahtonen is this theorem that presumably is proved in every basic course on Galois Theory: the compositum of two (and even infinitely many) abelian extensions is abelian. )
