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I believe I read somewhere that residually finite-by-$\mathbb{Z}$ groups are residually finite. That is, if $N$ is residually finite with $G/N\cong \mathbb{Z}$ then $G$ is residually finite.

However, I cannot remember where I read this, and nor can I find another place which says it. I was therefore wondering if someone could confirm whether this is true or not, and if it is give either a proof or a reference for this result? (If not, a counter-example would not go amiss!)

Note that I definitely know it is true if $N$ is f.g. free (this can be found in a paper of G. Baumslag, "Finitely generated cyclic extensions of free groups are residually finite" (Bull. Amer. Math. Soc., 5, 87-94, 1971)).

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1  
No; see my negative answer to this question: mathoverflow.net/questions/22814/… –  Tom Church Oct 17 '11 at 15:09
    
Thanks. However, it seems I have got my question upside-down. I meant to ask about residually finite-by-cyclic groups, not cyclic-by-residually finite...Sorry! –  user6503 Oct 17 '11 at 16:27

4 Answers 4

up vote 7 down vote accepted

The modified question has a positive answer if $N$ is finitely generated.

Consider an extension $1 \to N \to G \to \mathbb Z \to 1$ and take a lift $u \in G$ of the generator of $\mathbb Z$. If $N$ is finitely generated and $H' \subset N$ is a subgroup of finite index, then the intersection of all subgroups of index $[N:H']$ (call it $H$) is invariant under conjugation by $u$. Hence, for all $m \in \mathbb Z$, the subgroup $Hu^{m \mathbb Z}$ is a finite index normal subgroup of $G$.

Hence, if $N$ is finitely generated and residually finite, then $G$ is residually finite as well.

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(f.g. residually finite)-by-(residually finite) implies residually finite. It's an exercise and is well-known and can be found in many places (maybe Lyndon-Schupp but I can't check right now). –  YCor Oct 17 '11 at 18:25
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Yves - Deligne gave a counterexample to this. Or have I misunderstood you? In Andreas' answer, the point is that the extension splits. –  HJRW Oct 18 '11 at 9:15
    
Ah, very nice. Thanks. –  user6503 Oct 18 '11 at 9:53
    
yes sorry of course... I had the counterexamples in mind. As you point out, we need the extension be split (cf Osin's answer). Btw I think there are much easier examples than Deligne's (which relies on CSP and more) of finite-by-(residually finite) groups that are not residually finite. –  YCor Oct 21 '11 at 17:29

This is not true if $N$ is not assumed f.g. E.g. the wreath product of a nonabelian finite group $H$ by the integers is not residually finite (Gruenberg 1957, can be checked as a exercise). Here $N$ is an infinite direct sum of copies of $H$ (shifted by the action of the integers) and is residually finite.

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I think what Yves meant in his comment to Andreas' answer is the result of Mal'cev [A. I. Mal'cev, On homomorphisms onto finite groups, Ivanov. Gos. Ped. Inst. Ucen. Zap., 18 (1958), pp. 49-60, in Russian] stating that any split extension of a finitely generated residually finite group by a residually finite group is residually finite. This is indeed an easy exercise. In particular, this implies that every (f.g. residually finite) - by - Z group is residually finite as every extension by a free group (e.g., by Z) splits.

If either of the conditions "finitely generated" or "split" is relaxed, the result does not hold. The counterexample with infinitely generated kernel is already provided by Yves. For non-split extensions there are even examples of the form $$1\to \mathbb Z/2\mathbb Z \to G\to Q\to 1,$$ where $Q$ is (finitely generated) residually finite while $G$ is not. In particular, being residually finite is not a quasi-isometry invariant.

Such examples can be found among solvable (moreover, center-by-metabelian) groups or even among groups of intermediate growth (see [A. Erschler, Not residually finite groups of intermediate growth, commensurability and non-geometricity, Journal of Algebra, 272 (2004), 154-172].

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This is not true. The most prominent examples of non-residually finite central extensions of residually finite groups (by $\mathbb Z$) are certain lattices in non-linear Lie groups.

See for example

M. S. Raghunathan. Torsion in cocompact lattices in coverings of Spin(2, n). Math. Annalen 266, 403–419, 1984.

or

P. Deligne. Extensions centrales non residuellement finies de groupes arithmetiques. CR Acad. Sci. Paris, serie A-B, 287, 203–208, 1978.

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I apologise for my sloppyness - it seems I have got my question "upside-down", as it were. I meant to ask about residually finite-by-cyclic groups, not cyclic-by-residually finite! Sorry! (But thankyou for your answer, it is informative all the same.) –  user6503 Oct 17 '11 at 16:23
    
(I have edited the question appropriately now.) –  user6503 Oct 17 '11 at 16:26

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