Given a finite group $G$, I wish to show that the that product $\prod_{n \in \mathbb{N}} G$ is an amenable group.
It is known that for a group $H$, $H$ is amenable if and only if for all finite subset $S \subset H$ the subgroup $<S>$ of $H$ is amenable. So given $g_1,.., g_n \in \prod_{n \in \mathbb{N}} G$ I want to show that $<g_1,..., g_n>$ is amenable.
I was hoping to get to something by using the fact that since $G$ is finite we could say basically that all the information of $<g_1,..., g_n>$ is encoded somehow in the first $r$ coordinates for $r$ big enough since at some point all possible product have been done. However, I am not really sure it works nor how to do it cleanly. I was thinking of maybe getting an isomorphism to some direct sum which we know to be amenable or something like this.
So if someone could help me with this it would be appreciated. (Maybe there is an other approach more suited...)
It is amenable because it is locally finite.
That $G^X$ is locally finite when $G$ is finite (for every set $X$) is shown as follows.
First observe that a group $H$ is locally finite if and only if for every finitely generated free group $F$, every homomorphism $F\to H$ has a finite image. Consider a homomorphism $f=(f_x)_{x\in X}:F\to G^X$. Observe that $\mathrm{Hom}(F,G)\simeq G^d$ is finite. Then the kernel of $f$ is the intersection $\bigcap_x\mathrm{Ker}f_x$, which contains in the finite intersection $\bigcap_{f\in \mathrm{Hom}(F,G)}\mathrm{Ker}f$, which has finite index. Hence $f$ has a finite image.