To be more precise (but less snappy): is there an example of a group $G$ with finitely many infinite-index subgroups $H\_1,\dots, H\_n$ and elements $k\_1,\dots, k\_n$ such that $G$ is the union of the left cosets $k\_1 H\_1 , ..., k\_n H\_n\ ?$ And what if we relax the requirement that these all be *left* cosets, and ask: can $G$ be the union of finitely many such cosets, some being left cosets, others being right cosets?


If $G$ is [amenable](http://en.wikipedia.org/wiki/Amenable) then this can't happen, since any coset of an infinite-index subgroup must have measure $0$. So this immediately rules out any abelian group $G$.


I've tried playing around with the only non-amenable groups that I'm comfortable with, the free groups on two or more generators. A few months ago I thought I found a simple counterexample in the free group on $\aleph\_0$ generators, but now I've lost my notes and am beginning to doubt I ever had such an example.


(This question was asked to me by a friend who's interested in some kind of application to model theory, but I think it's interesting as a stand-alone puzzle.)



Can a group be a finite union of (left) cosets of infinite-index subgroups?
No. This follows from a beautiful theorem of B.H. Neumann:


Let $G$ be a group. If $\{x\_iH\_i\}\_{i=1}^n$ is a covering of $G$ by cosets of proper subgroups, then $n \geq \min\_{i} [G:H\_i]$.


Explicitly, this is Lemma 4.1 in


[http://alpha.math.uga.edu/~pete/Neumann54.pdf](http://alpha.math.uga.edu/%7Epete/Neumann54.pdf)


As Neumann remarks, the identity $gH = (g H g^{-1}) g$ shows that it is no loss of generality to restrict to coverings by left cosets.


