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Update: Steve points out in comments that many direct products of residually cyclic groups will be counterexamples. Indeed, I should have spotted this: every finitely generated abelian group is residually cyclic! I should have asked about fully residually cyclic groups, ie groups $G$ such that for any finite subset $X\subseteq G\smallsetminus 1$, there is a homomorphism from $G$ to a cyclic group that doesn't kill any elements of $X$. I suspect the question is now quite easy, though I haven't had a chance to think about it yet.

This question is motivated by this recent question, which asked for a characterisation of groups with at most one subgroup of each finite index. Arturo Magidin's answer showed that every such finite group is cyclic, but the questioner, Louis Burkill, added in comments that he is really interested in the infinite case.

In my answer to that question, I argue that a finitely generated group has at most one subgroup of each finite index if and only if its canonical residually finite quotient, $R(G)$, is cyclic. The `finitely generated' hypothesis is necessary - otherwise the additive group of the rationals provides a counterexample.

In the general case, one can reduce from the case of $G$ to $R(G)$ as before (this gets rid of pathological examples like infinite simple groups), and the same argument shows that if $R(G)$ has at most one subgroup of each finite index then $R(G)$ is fully residually cyclic, in particular abelian. But the converse is not clear to me. Hence the question in the title, which I'll reiterate here, with some extra hypotheses that should indicate where the difficulty lies.

If $G$ is an infinitely generated, fully residually cyclic (in particular, abelian) group, must $G$ have at most one subgroup of each finite index?

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Wouldn't the direct product of two residually cyclic groups be residually cyclic? – user641 Dec 1 '10 at 16:23
Um... yes. Good point. But not fully residually cyclic. Perhaps that's what's needed. – HJRW Dec 1 '10 at 17:24
@Paradoxial: Please stop bumping extremely old posts just for LaTeX. – Asaf Karagila Feb 13 '13 at 20:15
up vote 2 down vote accepted

The odds are against us. Even "fully" is not enough. With the "fully" hypothesis, probably it is easy: If $G$ has two subgroups $H,K$ of finite index $n$, then their intersection has finite index. Take $X$ to be a set of (non-identity) coset reps of $G/H \cap K$, then there is a quotient where no element non-identity element of $G/H \cap K$ is sent to the identity (and WLOG, G/H \cap K is that quotient), but since $G$ is fully residually cyclic, that quotient is cyclic, and the two subgroups must be identical by the lattice homomorphism theorem.

In other words, you seem to have exactly answered your own question! :)

At least K4 is no longer a counterexample.

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I think there's something wrong with this argument (and I suspect it's the WLOG part), because $\mathbb{Z}\times\mathbb{Z}$ is fully residually cyclic. – user641 Dec 1 '10 at 21:00
Yeah, Steve is right, and I was still wrong! $\mathbb{Z}^2$ is still fully residually cyclic. Which leaves me stumped. Is there a nice characterisation of the infinitely generated abelian groups with at most one subgroup of each finite index? – HJRW Dec 1 '10 at 22:56
Oh that's cool. K4=2×2 is not fully residually cyclic, right? Take G=Z×Z, H=Z×2Z and K=2Z×Z, then the specific quotient is not G/H∩K = K4. Instead, for X={(1,0),(0,1)} we get the quotient of K4 by its diagonal subgroup, that is, G/<(2,0),(0,2),(1,1)>. – Jack Schmidt Dec 1 '10 at 23:26
Actually X={(1,0),(0,1),(1,1)} and so I think the quotient has to be something like G/<(3,0),(1,-1)>. Then each element of X is nonzero in the quotient, but actually the first two elements become equal (and the last is their inverse). Super strange. – Jack Schmidt Dec 1 '10 at 23:40
It can also be something like G/<(1,2), (0,5)>. You can generally choose <(1,p)> to get ZxZ/ <(1,p)> ~ Z, such that no element of X is in the kernel, then use the residual finiteness of Z to finish. – user641 Dec 2 '10 at 0:16

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