How to use Markov-Kakutani fixed point theorem to show that abelian groups are amenable? Recall that a group $\Gamma$ is amenable if there exists a state $\mu$ on $l^{\infty}(\Gamma)$ which is invariant under the left translation action: for all $s\in \Gamma$ and $f\in l^{\infty}(\Gamma)$, $\mu(s.f)=\mu(f)$.
If the group is abelian, as the Markov-Kakutani fixed point theorem easily implies the group is amenable.
Well, my question is: what is Markov-Kakutani fixed point theorem here (I am no familiar with fixed point theory)? And how to use it to verify the abelian group is amenable?
 A: Let me first copy the proof from Ceccherini-Silberstein, Coornaert: Cellular Automata and Groups page 92 and then give some explanations. This is basically the same proof as the proof in this answer on MO.

Theorem 4.6.1. Every abelian group is amenable.
Proof. Let $G$ be an abelian group. Equip $(\ell^\infty(G))^*$ with the weak-* topology.
By Theorem 4.2.1, the set $\mathcal M(G)$ is a nonempty convex compact subset of $(\ell^\infty(G))^*$.
On the other hand, it follows from Proposition 4.3.1 that the action of $G$ on $\mathcal M(G)$ is affine and continuous (note that the left and the right actions coincide since $G$ is Abelian).
By applying the Markov-Kakutani fixed-point Theorem (Theorem G.1.1), we deduce that G has at least one fixed point in $\mathcal M(G)$. Such a fixed point is clearly a bi-invariant mean on $G$. This shows that $G$ is amenable.

Here $\mathcal M(G)$ denotes the set of all means on $G$, i.e., all positive functionals $\varphi \in (\ell^\infty)^*$ such that $\|\varphi\|=1$.
The above proof starts by the claim that this set is non-empty convex and compact.
It is nonempty, since for example the evaluation functional $\varphi(f)=f(e_G)$ is a mean on $G$. (It is the mean corresponding to the finitely additive measure $\mu$ defined by $\mu(A)=1$ if $e_G\in A$ and $\mu(A)=0$ otherwise.)
It is easy to see that convex combination of positive functionals is convex. Also showing that a unit ball of a linear normed space is convex is easy.
Compactness can be shown using Banach-Alaoglu theorem. This theorem says that the unit ball of $(\ell^\infty)^*$  with the weak-* topology is compact. Now it suffices to show that $\mathcal M(G)$ is closed subset of this unit ball. Perhaps the simplest approach is to show that a limit of any convergent net of means is again a mean.
Now we have a group action of $G$ on $\mathcal M(G)$ defined by $g\varphi(x)=\varphi(g^{-1}x)$. (You only need to verify, that that each $g\varphi$ is again in $\mathcal G$ and this is indeed a group action.)
Each of the maps $\varphi\mapsto g\varphi$ is a linear continuous map $\mathcal M(G)\to\mathcal M(G)$. Moreover, since the group $G$ is commutative, these maps commute with each other.
Markov-Kakutani fixed-point theorem says that if we have a system of maps from $C$ to $C$, such that each of these maps is affine and the maps commute with each other, where the set $C$ is non-empty compact convex subset of some Hausdorff topological vector space, then there exists a common fixed point of all these maps.
In this case all assumptions of this theorem are fulfilled, so there is a mean $\varphi\in\mathcal M(G)$, which is not changed any of the shifts $\varphi\mapsto\varphi g$, i.e. $g\varphi=\varphi$ for each $g\in G$. This means that $\varphi$ is an invariant mean on $G$.
