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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 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.)

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1 Answer 1

up vote 33 down vote accepted

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

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.

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That was quick! Thanks, actually I read about this a long time ago and forgot that it applied to any group G... –  John Goodrick Mar 7 '10 at 17:37
You're quite welcome. I have had a side interest in covering problems of various sorts for a couple of years now, and Neumann's result is one of the nicest in this area. –  Pete L. Clark Mar 7 '10 at 17:42
The link in this answer has decayed. Can someone give an updated link? Or, even better, a reference to the original paper of B. H. Neumann? –  Lee Mosher Sep 7 at 16:53
@Lee: I am sorry (more precisely, more annoyed than you) for the inconvenience. Please see my profile page for an explanation and a general fix (although unfortunately, no one seems to see this automatically). –  Pete L. Clark Sep 7 at 16:58
That works, thanks. –  Lee Mosher Sep 7 at 18:36

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