Question

To be more precise (but less snappy): is there an example of a group G with finitely many infinite-index subgroups H_1, ..., H_n and elements k_1, ..., 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 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.)

Answer 1

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://www.math.uga.edu/~pete/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.