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Recall that a group $G$ is residualy finite if for every non-zero element $g\in G$ there exists a homomorphism $\sigma:G\rightarrow H$ such that $H$ is finite and $\sigma(g)\neq 0$. Mal'cev's theorem says that if $k$ is a field then any finitely-generated subgroup of $GL_n(k)$ is residually finite. For a proof of this theorem, see Steve D (Smith?)s answer to this question MO:9628. Note that the theorem does not say that $GL_n(k)$ is residually finite.

Does anyone know any other classes of groups with the property that any finitely generated subgroup is residually finite.

I would prefer examples with the following two properties: first, the group is not itself residually finite, and second, it is not simply a subgroup of some $GL_n(k)$. So, this excludes, for instance, free groups.

Side question: is there a good name for this property?

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When one has a property of groups, and it holds for finitely-generated subgroups of a group, then the adjective "locally" is usually pinned on. So I would call your property "locally residually finite". –  Ian Agol Jul 9 '11 at 22:03
    
Ah, thanks. That's useful. –  Benjamin Antieau Jul 9 '11 at 22:36
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Stallings proves in his lecture notes that GL_n(R) is locally residually finite for any commutative ring R with unit. See math.berkeley.edu/~stall/math257 –  Benjamin Steinberg Jul 20 '11 at 14:38
    
Yes. This statement is Mal'cev's theorem. Thanks for the link to the notes though. Those look nice. –  Benjamin Antieau Jul 20 '11 at 18:31
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I think Malcev only proved this for the case of a field. The ring case is harder, –  Benjamin Steinberg Jul 30 '11 at 12:20
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4 Answers 4

If $G$ is a compact group, it has this property. This is because by Peter-Weyl compact groups are residually linear. Now use Malcev.

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Good example, although perhaps I should have said I wanted non-residually linear groups as well. –  Benjamin Antieau Jul 9 '11 at 22:35
    
So far, all listed examples are direct limits of residually linear groups. What are completely different examples? –  Benjamin Steinberg Jul 10 '11 at 2:35
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There are uncountably many Tarski monsters, which are finitely generated simple infinity groups whose proper subgroups are cyclic. No infinite simple group is residually finite, so Tarski monsters aren't, but their proper subgroups obviously are residually finite.

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I think OP wanted all fg subgroups to be residually finite, not just the proper ones, but I upvoted it anyway since I like Tarski monsters. –  Benjamin Steinberg Jul 10 '11 at 2:40
    
That's not quite what I was looking for, but it's really good to know about anyways. Thanks. –  Benjamin Antieau Jul 15 '11 at 17:45
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Perhaps a locally finite simple group would help? How about the subset of all even permutations of the natural numbers with finite support? Unless I am misremembering something, this should be a simple locally finite group that is not a subgroup of GL_n(k).

Gerhard "Email Me About System Design" Paseman, 2011.07.09

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Indeed locally finite groups, by definition, are locally residually finite. It is known that, for every prime $p$, there are uncountably many pairwise non-isomorphic locally finite groups which moreover are $p$-groups: see Burns, R. G. A wreath tower construction of countably infinite, locally finite groups. Math. Z. 105, 1968, 367–386 –  Alain Valette Jul 10 '11 at 20:52
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This is a subgroup of my example. –  Ian Agol Jul 10 '11 at 21:00
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There are simple examples, but probably not what you're looking for, such as a union $GL_{\infty}(k)=\cup_n GL_n(k)$, where $GL_n(k) \subset GL_{n+1}(k)$ embeds in the obvious way by thinking of $k^{n+1}=k^n \oplus k$.

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Yeah, I had thought of this too. –  Benjamin Antieau Jul 9 '11 at 21:41
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