## Looking for a Group

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- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Looking for a Group

I'm looking for a group with certain properties (or a proof that no such group exists), and I thought perhaps the minds of the xkcd math forum might be able to help! Here's what I'm interested in: A group G such that:

1) G is finitely generated

2) G is amenable (discrete definition, due to condition 1).

3) G has no surjection onto Z (that is, there is no group homomorphism f from G to the group of integers such that f is onto).

3a) Bonus points if G does not virtually surject onto Z, that is, G has no finite-index subgroup with a surjection onto Z.

4) G is residually finite, but not finite. (For those too lazy to click, residually finite means that for every element of the group, there's a homomorphism to a finite group that doesn't send the chosen element to the identity.)

That last condition may be equivalent to having finite-index subgroups of arbitrarily large index; I'm not sure. If you don't like definition 4, I'll still be fairly happy with a group satisfying the first three if it has finite-index subgroups of arbitrarily large index (that is, for any number n, G has a finite-index subgroup with index greater than n).

So for example, the group of integers satisfies properties 1, 2, and 4, but not 3. A direct sum of a countable number of finite groups satisfies properties 2, 3, and 4, but not 1. A finite group satisfies properties 1, 2, and 3, but not property 4. Does anyone have any ideas?

1) G is finitely generated

2) G is amenable (discrete definition, due to condition 1).

3) G has no surjection onto Z (that is, there is no group homomorphism f from G to the group of integers such that f is onto).

3a) Bonus points if G does not virtually surject onto Z, that is, G has no finite-index subgroup with a surjection onto Z.

4) G is residually finite, but not finite. (For those too lazy to click, residually finite means that for every element of the group, there's a homomorphism to a finite group that doesn't send the chosen element to the identity.)

That last condition may be equivalent to having finite-index subgroups of arbitrarily large index; I'm not sure. If you don't like definition 4, I'll still be fairly happy with a group satisfying the first three if it has finite-index subgroups of arbitrarily large index (that is, for any number n, G has a finite-index subgroup with index greater than n).

So for example, the group of integers satisfies properties 1, 2, and 4, but not 3. A direct sum of a countable number of finite groups satisfies properties 2, 3, and 4, but not 1. A finite group satisfies properties 1, 2, and 3, but not property 4. Does anyone have any ideas?

Last edited by MartianInvader on Sat Dec 13, 2008 2:36 pm UTC, edited 2 times in total.

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

### Re: Looking for a Group

As written, property 4) looks trivial; it seems to be very easy to satisfy if H = G, although I could be wrong.

I can't check amenability, but what about the free group on two generators?

I can't check amenability, but what about the free group on two generators?

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: Looking for a Group

Did I write 4) incorrectly? I think it's written right. If H = G, choose S to be any two-element set{a,b}. Then, let h = ab^{-1} , and then we have that hb = a, which is no good because hb shouldn't belong to S for any h. I think you misread my "for every h \in H" as a "there exists an h \in H". I'll edit it to make it a bit more clear.

There is a tiny mistake, I guess; I should have said for every non-identity h \in H, since clearly the identity sends S to itself.

Actually, I think condition 4) is equivalent to

4a) For any finite set S of G which does not contain the identity, there is a finite-index subgroup H disjoint from S.

The free group on two generators violates condition 3), for example, map it to \mathbb{Z} by sending each generator to 1, and you induce a surjective homomorphism. It also happens to not be amenable so it violates condition 2) as well. It does satisfy 1) and 4), however.

There is a tiny mistake, I guess; I should have said for every non-identity h \in H, since clearly the identity sends S to itself.

Actually, I think condition 4) is equivalent to

4a) For any finite set S of G which does not contain the identity, there is a finite-index subgroup H disjoint from S.

The free group on two generators violates condition 3), for example, map it to \mathbb{Z} by sending each generator to 1, and you induce a surjective homomorphism. It also happens to not be amenable so it violates condition 2) as well. It does satisfy 1) and 4), however.

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

### Re: Looking for a Group

MartianInvader wrote:4) G has the following property: For any finite subset (NOT subgroup) S of G, there is a finite-index subgroup H such that the action of H on G by left-multiplication cannot ever send an element of S to another element of S. That is to say, for all h \in H and s \in S, hs is NOT in S.

(snip)

A finite group satisfies properties 1, 2, and 3, but not property 4.

Either I'm reading this wrong or you oversimplified the statement of property 4. If G is a finite group, then the subgroup consisting of only the identity has finite index.

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5967**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

### Re: Looking for a Group

MartianInvader wrote:The free group on two generators violates condition 3)

Its also the simplest example of a non amenable group.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: Looking for a Group

Nitrodon wrote:MartianInvader wrote:4) G has the following property: For any finite subset (NOT subgroup) S of G, there is a finite-index subgroup H such that the action of H on G by left-multiplication cannot ever send an element of S to another element of S. That is to say, for all h \in H and s \in S, hs is NOT in S.

(snip)

A finite group satisfies properties 1, 2, and 3, but not property 4.

Either I'm reading this wrong or you oversimplified the statement of property 4. If G is a finite group, then the subgroup consisting of only the identity has finite index.

Whoops. Yeah, in the process of trying to make the statement as simple as possible, I lost the "and infinite" statement somewhere along the way. However, I realized that property 4 is equivalent to being residually finite. I'm going to edit the top post to reflect this.

I can't even find a group that satisfies 1), 3), and 4) If anyone has any ideas at all, I'd love to hear them. You don't need to give me a perfect proof; I'm just looking for ideas. Whether it's some group you know about that has weird properties and might work here, or just a nice source for counterexamples in group theory, I'd love to hear it. Thanks!

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

- eta oin shrdlu
**Posts:**451**Joined:**Sat Jan 19, 2008 4:25 am UTC

### Re: Looking for a Group

(Disclaimer: I don't know much about group theory, so this might be all wrong.) Property 3a seems to be just saying that G is pure torsion, i.e. that it has no free subgroups. With property 1 (finitely generated) and the requirement that G be infinite, you're looking for counterexamples to Burnside's problem. These include Golod-Shafarevich groups (about which I know only the name). Googling for "residually finite torsion amenable" I found this reference (PDF), which seems to describe what you're looking for (residually finite, finitely generated, torsion, amenable).MartianInvader wrote:I can't even find a group that satisfies 1), 3), and 4) If anyone has any ideas at all, I'd love to hear them. You don't need to give me a perfect proof; I'm just looking for ideas. Whether it's some group you know about that has weird properties and might work here, or just a nice source for counterexamples in group theory, I'd love to hear it. Thanks!

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: Looking for a Group

Thanks eta! That gives me just the type of group I wanted!

It's worth noting that property 3a isn't equivalent to being torsion. For example, the rational numbers under addition don't have any non-trivial mapping onto Z. (sketch proof: If a rational q maps to an integer n, what does q/2n map to?) I'm pretty sure there's finitely generated examples out there as well. However, torsion certainly implies 3a, so an amenable counterexample to Burnside's problem gives me what I'm looking for.

It's worth noting that property 3a isn't equivalent to being torsion. For example, the rational numbers under addition don't have any non-trivial mapping onto Z. (sketch proof: If a rational q maps to an integer n, what does q/2n map to?) I'm pretty sure there's finitely generated examples out there as well. However, torsion certainly implies 3a, so an amenable counterexample to Burnside's problem gives me what I'm looking for.

- eta oin shrdlu
**Posts:**451**Joined:**Sat Jan 19, 2008 4:25 am UTC

### Re: Looking for a Group

Ah, yes, I forgot the "finite index" part.MartianInvader wrote:Thanks eta! That gives me just the type of group I wanted!

It's worth noting that property 3a isn't equivalent to being torsion.

So out of curiosity, what are you going to do with this group?

### Re: Looking for a Group

eta oin shrdlu wrote:Ah, yes, I forgot the "finite index" part.MartianInvader wrote:Thanks eta! That gives me just the type of group I wanted!

It's worth noting that property 3a isn't equivalent to being torsion.

So out of curiosity, what are you going to do with this group?

Bet : He'll submit it to his professor for points! You did his homework, Eta!

Win : 9001 Internets!

Lose: I'll do your algebra for a week! (But I suck, so you really win nothing at all, nothing at all...)

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: Looking for a Group

No, it's not homework or anything. I was interested because I'm currently looking into something called property tau. There's no wikipedia entry, but it's related to property T. Basically, if a group surjects onto the integers, it has property tau. It's also the case that a group has property tau if it surjects onto a residually finite amenable group. I'm poking around to see if there's any cases where you could use the latter property when you couldn't use the former.

### Re: Looking for a Group

OK, this is getting quite complicated, so let's try to distill what's happened here:

The OP was looking for an infinite, finitely generated, amenable group, whose abelianization is finite, and is residually finite. More generally, he was looking for one of those groups such that all subgroups of finite index have finite abelianizations. (This property is known as FAb, short for 'finite abelianizations'.) There are plenty of examples of groups that are infinite, finitely generated, residually finite, and are FAb. Examples include the Nottingham groups, the various `index subgroups' of the Nottingham group found (independently) by Ben Klopsch and Fesenko, together with [imath]S[/imath]-arithmetic groups (e.g., [imath]\mathrm{PSL}_n(\mathbb{Z})[/imath]). (See a recent paper by Michael Larsen and Alex Lubotzky called 'representation growth for linear groups'.)

Now consider the (first) Grigorchuk group. This group is hereditarily just-infinite (I think, but check anyway), has intermediate growth (I know this) and is hence amenable, is residually finite (all groups acting as tree automorphisms are residually finite), and is 3-generator. Thus it is an example of a group satisfying all of your requirements.

If you only want the group to have finite abelianization, rather than having FAb, then examples include the infinite dihedral group (which is soluble, so certainly amenable), and the fours group, introduced by Don Passman. This group is an interesting example of a virtually abelian, torsion-free group, which is not right-orderable (and has other properties that people care about). It is given by[math]\Gamma=\langle x,y\mid x^{-1}y^2x=y^{-2},\;y^{-1}x^2y=x^{-2}\rangle,[/math]with the subgroup generated by [imath]x^2[/imath], [imath]y^2[/imath] and [imath](xy)^2[/imath] being an abelian normal subgroup. (This group has a homocyclic abelianization of type (4,4).)

There are interesting questions concerning these FAb groups, particularly in the emergent field of `representation growth', which is still in its infancy. I'm in the process of producing a preprint on representation growth, and so I know something about such groups!

The OP was looking for an infinite, finitely generated, amenable group, whose abelianization is finite, and is residually finite. More generally, he was looking for one of those groups such that all subgroups of finite index have finite abelianizations. (This property is known as FAb, short for 'finite abelianizations'.) There are plenty of examples of groups that are infinite, finitely generated, residually finite, and are FAb. Examples include the Nottingham groups, the various `index subgroups' of the Nottingham group found (independently) by Ben Klopsch and Fesenko, together with [imath]S[/imath]-arithmetic groups (e.g., [imath]\mathrm{PSL}_n(\mathbb{Z})[/imath]). (See a recent paper by Michael Larsen and Alex Lubotzky called 'representation growth for linear groups'.)

Now consider the (first) Grigorchuk group. This group is hereditarily just-infinite (I think, but check anyway), has intermediate growth (I know this) and is hence amenable, is residually finite (all groups acting as tree automorphisms are residually finite), and is 3-generator. Thus it is an example of a group satisfying all of your requirements.

If you only want the group to have finite abelianization, rather than having FAb, then examples include the infinite dihedral group (which is soluble, so certainly amenable), and the fours group, introduced by Don Passman. This group is an interesting example of a virtually abelian, torsion-free group, which is not right-orderable (and has other properties that people care about). It is given by[math]\Gamma=\langle x,y\mid x^{-1}y^2x=y^{-2},\;y^{-1}x^2y=x^{-2}\rangle,[/math]with the subgroup generated by [imath]x^2[/imath], [imath]y^2[/imath] and [imath](xy)^2[/imath] being an abelian normal subgroup. (This group has a homocyclic abelianization of type (4,4).)

There are interesting questions concerning these FAb groups, particularly in the emergent field of `representation growth', which is still in its infancy. I'm in the process of producing a preprint on representation growth, and so I know something about such groups!

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