I’m collecting in this blog post a number of simple group-theoretic lemmas, all of the following flavour: if is a subgroup of some product of groups, then one of three things has to happen:

- ( too small) is contained in some proper subgroup of , or the elements of are constrained to some sort of equation that the full group does not satisfy.
- ( too large) contains some non-trivial normal subgroup of , and as such actually arises by pullback from some subgroup of the quotient group .
- (Structure) There is some useful structural relationship between and the groups .

It is perhaps easiest to explain the flavour of these lemmas with some simple examples, starting with the case where we are just considering subgroups of a single group .

Lemma 1Let be a subgroup of a group . Then exactly one of the following hold:

- (i) ( too small) There exists a non-trivial group homomorphism into a group such that for all .
- (ii) ( normally generates ) is generated as a group by the conjugates of .

*Proof:* Let be the group normally generated by , that is to say the group generated by the conjugates of . This is a normal subgroup of containing (indeed it is the smallest such normal subgroup). If is all of we are in option (ii); otherwise we can take to be the quotient group and to be the quotient map. Finally, if (i) holds, then all of the conjugates of lie in the kernel of , and so (ii) cannot hold.

Here is a “dual” to the above lemma:

Lemma 2Let be a subgroup of a group . Then exactly one of the following hold:

- (i) ( too large) is the pullback of some subgroup of for some non-trivial normal subgroup of , where is the quotient map.
- (ii) ( is core-free) does not contain any non-trivial conjugacy class .

*Proof:* Let be the normal core of , that is to say the intersection of all the conjugates of . This is the largest normal subgroup of that is contained in . If is non-trivial, we can quotient it out and end up with option (i). If instead is trivial, then there is no non-trivial element that lies in the core, hence no non-trivial conjugacy class lies in and we are in option (ii). Finally, if (i) holds, then every conjugacy class of an element of is contained in and hence in , so (ii) cannot hold.

For subgroups of nilpotent groups, we have a nice dichotomy that detects properness of a subgroup through abelian representations:

Lemma 3Let be a subgroup of a nilpotent group . Then exactly one of the following hold:

- (i) ( too small) There exists non-trivial group homomorphism into an abelian group such that for all .
- (ii) .

Informally: if is a variable ranging in a subgroup of a nilpotent group , then either is unconstrained (in the sense that it really ranges in all of ), or it obeys some abelian constraint .

*Proof:* By definition of nilpotency, the lower central series

Since is a normal subgroup of , is also a subgroup of . Suppose first that is a proper subgroup of , then the quotient map is a non-trivial homomorphism to an abelian group that annihilates , and we are in option (i). Thus we may assume that , and thus

Note that modulo the normal group , commutes with , hence and thus We conclude that . One can continue this argument by induction to show that for every ; taking large enough we end up in option (ii). Finally, it is clear that (i) and (ii) cannot both hold.

Remark 4When the group is locally compact and is closed, one can take the homomorphism in Lemma 3 to be continuous, and by using Pontryagin duality one can also take the target group to be the unit circle . Thus is now a character of . Similar considerations hold for some of the later lemmas in this post. Discrete versions of this above lemma, in which the group is replaced by some orbit of a polynomial map on a nilmanifold, were obtained by Leibman and are important in the equidistribution theory of nilmanifolds; see this paper of Ben Green and myself for further discussion.

Here is an analogue of Lemma 3 for special linear groups, due to Serre (IV-23):

Lemma 5Let be a prime, and let be a closed subgroup of , where is the ring of -adic integers. Then exactly one of the following hold:

- (i) ( too small) There exists a proper subgroup of such that for all .
- (ii) .

*Proof:* It is a standard fact that the reduction of mod is , hence (i) and (ii) cannot both hold.

Suppose that (i) fails, then for every there exists such that , which we write as

We now claim inductively that for any and , there exists with ; taking limits as using the closed nature of will then place us in option (ii).The case is already handled, so now suppose . If , we see from the case that we can write where and . Thus to establish the claim it suffices to do so under the additional hypothesis that .

First suppose that for some with . By the case, we can find of the form for some . Raising to the power and using and , we note that

giving the claim in this case.Any matrix of trace zero with coefficients in is a linear combination of , , and is thus a sum of matrices that square to zero. Hence, if is of the form , then for some matrix of trace zero, and thus one can write (up to errors) as the finite product of matrices of the form with . By the previous arguments, such a matrix lies in up to errors, and hence does also. This completes the proof of the case.

Now suppose and the claim has already been proven for . Arguing as before, it suffices to close the induction under the additional hypothesis that , thus we may write . By induction hypothesis, we may find with . But then , and we are done.

We note a generalisation of Lemma 3 that involves two groups rather than just one:

Lemma 6Let be a subgroup of a product of two nilpotent groups . Then exactly one of the following hold:

- (i) ( too small) There exists group homomorphisms , into an abelian group , with non-trivial, such that for all , where is the projection of to .
- (ii) for some subgroup of .

*Proof:* Consider the group . This is a subgroup of . If it is all of , then must be a Cartesian product and option (ii) holds. So suppose that this group is a proper subgroup of . Applying Lemma 3, we obtain a non-trivial group homomorphism into an abelian group such that whenever . For any in the projection of to , there is thus a unique quantity such that whenever . One easily checks that is a homomorphism, so we are in option (i).

Finally, it is clear that (i) and (ii) cannot both hold, since (i) places a non-trivial constraint on the second component of an element of for any fixed choice of .

We also note a similar variant of Lemma 5, which is Lemme 10 of this paper of Serre:

Lemma 7Let be a prime, and let be a closed subgroup of . Then exactly one of the following hold:

- (i) ( too small) There exists a proper subgroup of such that for all .
- (ii) .

*Proof:* As in the proof of Lemma 5, (i) and (ii) cannot both hold. Suppose that (i) does not hold, then for any there exists such that . Similarly, there exists with . Taking commutators of and , we can find with . Continuing to take commutators with and extracting a limit (using compactness and the closed nature of ), we can find with . Thus, the closed subgroup of does not obey conclusion (i) of Lemma 5, and must therefore obey conclusion (ii); that is to say, contains . Similarly contains ; multiplying, we end up in conclusion (ii).

The most famous result of this type is of course the Goursat lemma, which we phrase here in a somewhat idiosyncratic manner to conform to the pattern of the other lemmas in this post:

Lemma 8 (Goursat lemma)Let be a subgroup of a product of two groups . Then one of the following hold:

- (i) ( too small) is contained in for some subgroups , of respectively, with either or (or both).
- (ii) ( too large) There exist normal subgroups of respectively, not both trivial, such that arises from a subgroup of , where is the quotient map.
- (iii) (Isomorphism) There is a group isomorphism such that is the graph of . In particular, and are isomorphic.

Here we almost have a trichotomy, because option (iii) is incompatible with both option (i) and option (ii). However, it is possible for options (i) and (ii) to simultaneously hold.

*Proof:* If either of the projections , from to the factor groups (thus and fail to be surjective, then we are in option (i). Thus we may assume that these maps are surjective.

Next, if either of the maps , fail to be injective, then at least one of the kernels , is non-trivial. We can then descend down to the quotient and end up in option (ii).

The only remaining case is when the group homomorphisms are both bijections, hence are group isomorphisms. If we set we end up in case (iii).

We can combine the Goursat lemma with Lemma 3 to obtain a variant:

Corollary 9 (Nilpotent Goursat lemma)Let be a subgroup of a product of two nilpotent groups . Then one of the following hold:

- (i) ( too small) There exists and a non-trivial group homomorphism such that for all .
- (ii) ( too large) There exist normal subgroups of respectively, not both trivial, such that arises from a subgroup of .
- (iii) (Isomorphism) There is a group isomorphism such that is the graph of . In particular, and are isomorphic.

*Proof:* If Lemma 8(i) holds, then by applying Lemma 3 we arrive at our current option (i). The other options are unchanged from Lemma 8, giving the claim.

Now we present a lemma involving three groups that is known in ergodic theory contexts as the “Furstenberg-Weiss argument”, as an argument of this type arose in this paper of Furstenberg and Weiss, though perhaps it also implicitly appears in other contexts also. It has the remarkable feature of being able to enforce the abelian nature of one of the groups once the other options of the lemma are excluded.

Lemma 10 (Furstenberg-Weiss lemma)Let be a subgroup of a product of three groups . Then one of the following hold:

- (i) ( too small) There is some proper subgroup of and some such that whenever and .
- (ii) ( too large) There exists a non-trivial normal subgroup of with abelian, such that arises from a subgroup of , where is the quotient map.
- (iii) is abelian.

*Proof:* If the group is a proper subgroup of , then we are in option (i) (with ), so we may assume that

As before, we can combine this with previous lemmas to obtain a variant in the nilpotent case:

Lemma 11 (Nilpotent Furstenberg-Weiss lemma)Let be a subgroup of a product of three nilpotent groups . Then one of the following hold:

- (i) ( too small) There exists and group homomorphisms , for some abelian group , with non-trivial, such that whenever , where is the projection of to .
- (ii) ( too large) There exists a non-trivial normal subgroup of , such that arises from a subgroup of .
- (iii) is abelian.

Informally, this lemma asserts that if is a variable ranging in some subgroup , then either (i) there is a non-trivial abelian equation that constrains in terms of either or ; (ii) is not fully determined by and ; or (iii) is abelian.

*Proof:* Applying Lemma 10, we are already done if conclusions (ii) or (iii) of that lemma hold, so suppose instead that conclusion (i) holds for say . Then the group is not of the form , since it only contains those with . Applying Lemma 6, we obtain group homomorphisms , into an abelian group , with non-trivial, such that whenever , placing us in option (i).

The Furstenberg-Weiss argument is often used (though not precisely in this form) to establish that certain key structure groups arising in ergodic theory are abelian; see for instance Proposition 6.3(1) of this paper of Host and Kra for an example.

One can get more structural control on in the Furstenberg-Weiss lemma in option (iii) if one also broadens options (i) and (ii):

Lemma 12 (Variant of Furstenberg-Weiss lemma)Let be a subgroup of a product of three groups . Then one of the following hold:

- (i) ( too small) There is some proper subgroup of for some such that whenever . (In other words, the projection of to is not surjective.)
- (ii) ( too large) There exists a normal of respectively, not all trivial, such that arises from a subgroup of , where is the quotient map.
- (iii) are abelian and isomorphic. Furthermore, there exist isomorphisms , , to an abelian group such that

The ability to encode an abelian additive relation in terms of group-theoretic properties is vaguely reminiscent of the group configuration theorem.

*Proof:* We apply Lemma 10. Option (i) of that lemma implies option (i) of the current lemma, and similarly for option (ii), so we may assume without loss of generality that is abelian. By permuting we may also assume that are abelian, and will use additive notation for these groups.

We may assume that the projections of to and are surjective, else we are in option (i). The group is then a normal subgroup of ; we may assume it is trivial, otherwise we can quotient it out and be in option (ii). Thus can be expressed as a graph for some map . As is a group, must be a homomorphism, and we can write it as for some homomorphisms , . Thus elements of obey the constraint .

If or fails to be injective, then we can quotient out by their kernels and end up in option (ii). If fails to be surjective, then the projection of to also fails to be surjective (since for , is now constrained to lie in the range of ) and we are in option (i). Similarly if fails to be surjective. Thus we may assume that the homomorphisms are bijective and thus group isomorphisms. Setting to the identity, we arrive at option (iii).

Combining this lemma with Lemma 3, we obtain a nilpotent version:

Corollary 13 (Variant of nilpotent Furstenberg-Weiss lemma)Let be a subgroup of a product of three groups . Then one of the following hold:

- (i) ( too small) There are homomorphisms , to some abelian group for some , with not both trivial, such that whenever .
- (ii) ( too large) There exists a normal of respectively, not all trivial, such that arises from a subgroup of , where is the quotient map.
- (iii) are abelian and isomorphic. Furthermore, there exist isomorphisms , , to an abelian group such that

Here is another variant of the Furstenberg-Weiss lemma, attributed to Serre by Ribet (see Lemma 3.3):

Lemma 14 (Serre’s lemma)Let be a subgroup of a finite product of groups with . Then one of the following hold:

- (i) ( too small) There is some proper subgroup of for some such that whenever .
- (ii) ( too large) One has .
- (iii) One of the has a non-trivial abelian quotient .

*Proof:* The claim is trivial for (and we don’t need (iii) in this case), so suppose that . We can assume that each is a perfect group, , otherwise we can quotient out by the commutator and arrive in option (iii). Similarly, we may assume that all the projections of to , are surjective, otherwise we are in option (i).

We now claim that for any and any , one can find with for and . For this follows from the surjectivity of the projection of to . Now suppose inductively that and the claim has already been proven for . Since is perfect, it suffices to establish this claim for of the form for some . By induction hypothesis, we can find with for and . By surjectivity of the projection of to , one can find with and . Taking commutators of these two elements, we obtain the claim.

Setting , we conclude that contains . Similarly for permutations. Multiplying these together we see that contains all of , and we are in option (ii).

## 18 comments

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8 May, 2021 at 6:03 am

Ehud SchreiberHi,

it seems a prime is missing in Lemma 8 (i), where it shows

h_3 \in G_3

Thanks,

Ehud.

[Corrected, thanks – T.]8 May, 2021 at 7:34 am

Emmanuel KowalskiGoursat’s lemma and the generalizations to more than two factors (and variants involving Lie algebras) arise fairly frequently in arithmetic geometry.

The earliest reference I know is Serre’s 1972 paper “Propriétés galoisiennes des points d’ordre fini des courbes elliptiques” (see section 6 there), then there are papers of Ribet and others. Katz uses extensively such ideas under the name “Goursat-Kolchin-Ribet criterion” (in his book “Exponential sums and differential equations” for instance), but I never tracked Kolchin’s contribution.

It’s one of the key underlying mechanisms in estimates for multiple correlations of translated Kloosterman sums, among many similar applications to analytic number theory.

I didn’t know the Furstenberg-Weiss reference, but the argument (with the commutator trick) does seem related to Lemma 3.3 in Ribet’s 1975 paper “On ell-adic representations attached to modular forms” (who attributes the proof to Serre); this is then used to get statements involving finite products with any number of factors.

In (maybe) similar spirit, number theory often uses statements like “A subgroup of SL_2(Z_p) which surjects to SL_2(F_p) is equal to SL_2(Z_p)” (which is true for p at least 5, if I remember correctly) to reduce problems about profinite groups to the case of finite groups. This also generalizes to other groups, in suitable form.

9 May, 2021 at 10:18 am

Terence TaoThanks for these references! I have incorporated these results into the text.

8 May, 2021 at 11:09 am

PShould the first instance of “or obeys” be there?

[Corrected, thanks – T.]9 May, 2021 at 11:36 am

Paul F. Dietz (@PaulFDietz)“Combining this lemma with Lemma \reF{nilp}, we obtain a nilpotent version:”

Typo

[Corrected, thanks – T.]11 May, 2021 at 12:02 pm

CrystallineIn the proof of Lemma 5, there is a typo. “Suppose that (i) fails, then for every g \in SL_2(\ZZ_p) there exists h \in SL_2(\ZZ_p)” , you mean h \in H.

Also, when you deal with the case g = 1+pX+O(p^2), you note that one can find an h of the form 1+X+pY+O(p^2), and you then say that, raising to the pth power and using X^2=0 and p \geq 5, we get that h^p = 1+ pX +O(p^2).

I managed to show this by explicitly writing down a recursive formula for the coefficients of h^n, n \geq 3. It wasn’t hard, but there was still some computations to be made, so I wanted to ask, is there a direct way to see that h^p = 1+pX+O(p^2), without writing down a formula for the coefficients?

[Corrected and expanded, thanks – T.]12 May, 2021 at 6:31 am

YahyaAA1In the proof of Lemma 5,there’s a repeated word: “… such such …”.

[Corrected, thanks – T.]13 May, 2021 at 4:47 pm

Goursat and Furstenberg-Weiss type lemmas – scroo0ooge[…] the group is locally compact and is closed, one can take the homomorphism in Lemma 3 to be continuous, and by using Pontryagin duality one can also take the target group to be the […]

14 May, 2021 at 3:04 am

OttoA minor typo founded at the end of the proof of lemma 12: “similarly” instead of “simiarly”.

[Corrected, thanks – T.]14 May, 2021 at 4:21 pm

SamIn the proof of lemma 6 : “One easily checcks that” “One easily checks that”

[Corrected, thanks – T.]19 May, 2021 at 10:52 pm

AnonymousDear Pro.Tao,

Recently, I have heard of four mathematicians and even AI programs solving Navier Stokes equation.

Today, I have some advice:

+ Navier Stokes problem is water or other liquid problem.

+ Firestly, I’d like to say that nature itself is answer and number.For example, a day has 24 hours, a year has 12 months , a year has 4 seasons, ….and all human beings were born after 9 months and 10 days. You notice that these numbers are even number.

+ We know that water boils at 100°C.

+ I guarantee that “finite time blow up Navier Stokes equations that you concern is even number.

+And today, I’d like to conclude that great and super science without philosophy will never be successful.

+Why do all mathematicians in the world keep their heads solving Navier Stokes with one way?

+ Likely, you close a screw, you must turn it from left to right.When you open it , you must anticlockwise(from right to left).

Navier Stokes is water.We must find something is opposite it. That’s fire!

+Water is plentiful of shapes and fire is also like

+ Water can rise object up and fire can also like this.

+ You imagine you have an ancient oil lamp. First, its flame is very little and then 1 minute, it blow up a big flame. Movement of fire is also like movement of water.

+Hundreds of years, all scientists in the world only research of water, no one research of fire.

+ Navier Stokes invented water equation. Why do not you invent fire equation and solve it? Secrete of water is also secrete of fire.( Fire can turn into water).

+About this, you should use Topology knowledge to solve it.

+Navier Stokes itself is looking for you, you can not choose it although you are genuis . Navier Stokes is very suitable for you.

+Likely, a man wins Jackpot lotterly ( itself numbers choose its owner, no one can choose number)

+Today, I confirm that your Navier Stokes work is absolutely right.I say the truth. You never be doubtful of your work . You try little, you are the only mathematician in the world being successfull in solving Navier Stokes in this year.

+ wake up and go! Continue! A ripe fruit is waiting for you. This is your duty and mission.!

* You do not need to know me who is . You close your eyes, you will see and understand. I always help you. I am not a mathematician , I am a superman in other field.

Best wishes,

28 May, 2021 at 1:00 pm

AnonymousWise words.

23 May, 2021 at 3:33 am

Michael VilloriaHi, Professor Tao! I’m a huge fan, keep up the great work!

25 May, 2021 at 12:34 pm

valuevar“Next, if either of the maps , fail to be injective, then at least one of the kernels , is non-trivial. We can then descend down to the quotient and end up in option (ii).”

Isn’t it exactly the other way around? is a subgroup of of the form for some ; is a subgroup of of the form for some . Then we can descend down to the quotient and end up in option (ii).

[Corrected, thanks – T.]27 May, 2021 at 6:01 am

Sabino LamonacaVery nice work prof Tao.

28 May, 2021 at 10:12 am

SteveThere is a typo in the (ii) of Lemma 12: a comma is missing between G2 and G3 i.e. “of G1, G2, G3 respectively”

[Corrected, thanks – T.]29 May, 2021 at 8:20 am

SteveI my previous post I have forgotten to point out that the same typo occurs in the Corollary 13 (ii) (H too large): “of G1, G2, G3”

[Corrected, thanks – T.]19 September, 2021 at 12:20 pm

Pierre de la HarpeGoursat lemma (= Lemma 8), often stated in a slighlty more precise form,

has appeared several times and has been forgotten several times.

Some references and the standard proof in the attached pdf.

/Users/laharpe/Desktop/GoursatL19sept21.pdf