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arXiv:1109.0024v2 [math.GR] 19 Sep 2011

A GENERALIZED GOURSAT LEMMA

K. Bauer, D. Sen, P. Zvengrowski

Abstract

In this note the usual Goursat lemma, which describes subgroups

of the direct product of two groups, is generalized to describing sub-

groups of a direct product A1× A2× ··· × An of a finite number of

groups. Other possible generalizations are discussed and an application

to cyclic subgroups is given.

1 Introduction

In Sections 11-12 of a paper written in 1889 [Gou89], the famed French math-

ematician´Edouard Goursat developed what is now called Goursat’s lemma

(also called Goursat’s theorem or Goursat’s other theorem), for characteriz-

ing the subgroups of the direct product A×B of two groups A,B. It seems

to have been first attributed to Goursat by J. Lambek in [Lam58, Lam76],

who in turn attributes H.S.M. Coxeter for bringing this to his attention.

The lemma is elementary and a natural question to consider, for example

it appears as Exercise 5, p. 75, in Lang’s Algebra [Lan02]. It has also been

the subject of recent expository articles [Pet09, Pet11] in an undergraduate

mathematics journal. It is possible that other authors discovered the lemma

independently without knowing the original reference. Indeed, one such ex-

ample occurs in 1961 in a paper of A. Hattori, ([Hat61], Section 2.3), now

translated into English [HZ09] from the original Japanese.

Other sources that mention Goursat’s lemma include papers of S. Dick-

son [Dic69] in 1969, K. Ribet [Rib76] in 1976, a book by R. Schmidt ([Sch94],

Chapter 1.6), a paper of D. Anderson and V. Camillo in 2009 [AC09], and

several internet sites such as [FL10, AEM09].

12010 Mathematics Subject Classification : 20, 18.

Keywords and phrases : Goursat’s Lemma

2The third named author was supported during this work by a Discovery Grant from

the Natural Sciences and Engineering Research Council of Canada.

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There are a number of interesting possibilities for generalizing this useful

lemma. The first is to subgroups of a semi-direct product, and this is stud-

ied in [Use91]. The second is to other categories besides groups. Indeed,

it is proved for modules in [Lam76], and this implies that it will hold in

any abelian category by applying the embedding theorems of Lubkin-Freyd-

Heron-Mitchell (cf. [Mac71], p. 205). It is proved for rings in [AC09]. The

most general category in which one can hope to have a Goursat lemma is

likely an exact Mal’cev category, cf. [FL10], although to the best of our

knowledge a a proof of this fact is yet to be given.

In this note we examine another generalization, to the direct product of

a finite number of groups. While this sounds at first glance to be a triviality

since we can write A×B×C ≈ (A×B)×C, unexpected complications arise

as noted by Arroyo et al. in [AEM09]. The complications are overcome by

introducing an asymmetric version of the lemma in Section 2, which enables

us to solve the general case in Section 3. Applications within group theory,

in particular to cyclic subgroups of a direct product A1×···×An, are given

in Section 4.

2 An asymmetric version of Goursat’s lemma

For convenience, we start by stating the usual version of Goursat’s lemma.

Let A,B be groups and G ⊆ A × B be a subgroup. The neutral element of

each group A and B, with slight abuse of notation, will be written ‘e’. Let

π1: A × B → A, π2: A × B → B be the natural projections and ı1: A →

A × B, ı2: B → A × B be the usual inclusions.

2.1 Theorem (Goursat’s lemma) There is a bijective correspondence be-

tween subgroups G of A × B and quintuples {G1,G1,G2,G2,θ}, where

G1✁ G1⊆ A, G2✁ G2⊆ B, and θ: G1/G1

≈

− → G2/G2is an isomorphism.

As mentioned above, the proof is elementary and given as an exercise in

[Lan02], it can also be found in [AC09, Hat61]. The basic idea of the proof

is as follows. Suppose that G is a subgroup of A × B. Write G1= π1(G) =

{a ∈ A|(a,b) ∈ G for some b ∈ B}, G1 = ı−1

similarly for G2 and G2. It is easily seen that G1✁ G1, G2✁ G2. The

isomorphism θ: G1/G1

− → G2/G2is given by θ[a] = [b], where (a,b) ∈ G

and [a] = G1a, [b] = G2b are the respective cosets of a and b in G1/G1,

G2/G2(again with slight abuse of notation). Thus G determines a quintuple

Q′

2(G) = {G1,G1,G2,G2,θ}.

1(G) = {a ∈ A|(a,e) ∈ G},

≈

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Conversely, given a quintuple Q′= {G1,G1,G2,G2,θ}, let Γ′

the subgroup p−1(Gθ), where Gθ⊆ G1/G1× G2/G2is the graph of θ and

p: G1× G2→ G1/G1× G2/G2is the natural surjection. The functions Q′

and Γ′

2are inverse to each other.

We now state an equivalent asymmetric version of the lemma, which is

in effect just a minor variation of Theorem 2.1 but has the advantage that

it generalizes easily to higher direct products, as we shall see in Section 3.

2(Q′) be

2

2.2 Theorem (Asymmetric version of Goursat’s lemma) There is a

bijective correspondence between subgroups G of A × B and quadruples

{G1,G2,G2,θ1}, where G1⊆ A, G2✁ G2⊆ B are arbitrary subgroups of A

and B, and θ1: G1։ G2/G2is a surjective homomorphism.

Proof.

A × B we define Q2(G) to be the quadruple

The proof is similar to that of Theorem 2.1. For any subgroup G of

{G1,G2,G2,θ1},

where G1, G2and G2are the first, third and fourth coordinates of Q′

(cf. Theorem 2.1). The surjection θ1is given by θ1(a) = [b] for a ∈ G1and

(a,b) ∈ G for some b ∈ G2.

Conversely, for an arbitrary quadruple Q = {G1,G2,G2,θ1}, with G1⊆

A, G2✁ G2⊆ B, and θ1: G1։ G2/G2a surjective homomorphism define

2(G)

Γ2(Q) := p−1(Gθ1),

where Gθ1⊆ G1× G2/G2is the graph of θ1and p: G1× G2→ G1× G2/G2

is the natural surjection. The functions Q2and Γ2are inverse to each other.

?

The equivalence of Theorem 2.1 and Theorem 2.2 is easily seen. It need

only be pointed out that θ determines the surjection θ1as the composition

G1։ G1/G1

θ

≈G2/G2,

− →

while θ1determines θ via the first isomorphism theorem, specifically

G1

G2/G2

????????

???

???

????

G1/Ker(θ1)

??

θ1

??????????????

≈

θ

.

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Finally, we have

Ker(θ1) = {a ∈ G1|θ1(a) = [b],with b ∈ G2}

= {a ∈ G1|(a,b) ∈ G, (e,b) ∈ G}

= {a ∈ G1|(a,e) ∈ G}

= G1.

2.3 Remark It is often useful to note the additional fact that (G1×G2)✁G

and G/(G1× G2) ≈ G1/G1≈ G2/G2[Hat61], [HZ09].

3 The generalized Goursat lemma

As mentioned in the Introduction, generalizing the Goursat lemma from

n = 2 to n ≥ 2 seems to create unexpected complications. However, using

the asymmetric version Theorem 2.2, the generalization to finite n ≥ 2

becomes routine. We will next state the result for n = 3, after introducing

some convenient notation for any subgroup G ⊆ A1× ··· × An.

3.1 Definition Let S ⊂ {1,2,··· ,n} = n, and j ∈ n ? S. Then

G(j|S) := {xj∈ Aj|(x1,··· ,xj,··· ,xn) ∈ G

for some x1,··· ,xj−1,xj+1,··· ,xnwith xi= e if i ∈ S}.

For example,

G(1|∅) = π1(G), G(1|{2,3,··· ,n − 1}) = {x1∈ A1|(x1,e,··· ,e)} ∈ G}.

These correspond to the notation used in Section 2, when n = 2, via

G(1|∅) = G1, G(1|{2}) = G1, G(2|∅) = G2, G(2|{1}) = G2. For conve-

nience, we shall usually omit the brackets {}, e.g. G(1|{2,3}) = G(1|2,3).

Note that if T ⊆ S, then G(j|S) ✁ G(j|T).

3.2 Definition With S as above, we denote by πSthe standard projection

A1× ··· × An։?

··· × An. Then we set

j∈SAjand ıSthe standard inclusion?

j∈SAj֒→ A1×

GS:= πS(G) ⊆

?

j∈S

Aj,GS:= ı−1

S(G) ⊆

?

j∈S

Aj.

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For example, if n = 3, A1= A, A2= B, A3= C, and S = {1,3}, then

GS= {(a,c)|(a,b,c) ∈ G for some b ∈ B} and GS= {(a,c)|(a,e,c) ∈ G}. In

case S = {j}, GS= G{j}is written Gj, which is consistent with the notation

in Section 2, and similarly GS = G{j}= Gj = {aj ∈ Aj|(a1,...,an) ∈

G with ai = e,i ?= j}. One has GS✁ GS ⊆ A × C. We now state the

generalized Goursat lemma for n = 3, the extension for n ≥ 3 should be

clear (cf. Remark 3.4).

3.3 Theorem (Goursat’s lemma for n = 3) There is a bijective corre-

spondence between the subgroups of A × B × C and 7-tuples

{G1,G2,G(2|1),θ1,G3,G(3|1,2),θ2},

where G1⊆ A, G(2|1) ✁G2⊆ B, G(3|1,2) ✁G3⊆ C, θ1: G1։ G2/G(2|1),

and θ2: Λ ։ G3/G(3|1,2) are surjective homomorphisms with

Λ = Γ2({G1,G2,G(2|1),θ1}) ⊆ A × B

(cf. Theorem 2.2 for the definition of Γ2).

Proof.

If G is a subgroup of A × B × C, define the 7-tuple

Q3(G) = {G1,G2,G(2|1),θ1,G3,G(3|1,2),θ2}

where G1, G2, G(2|1), G3, and G(3|1,2) are the groups of Definitions 3.1

and 3.2. Further, for a ∈ G1we have θ1(a) = [b], where (a,b,c) ∈ G for some

b ∈ B,c ∈ C , and for (a,b) ∈ Λ, we have θ2(a,b) = [c] where (a,b,c) ∈ G

for some c ∈ C.

Conversely, given any 7-tuple

Q = {G1,G2,G(2|1),θ1,G3,G(3|1,2),θ2},

satisfying the hypotheses of Theorem 3.3, a subgroup Γ3(Q) ⊆ A × B × C

is determined by Γ3(Q) = p−1(Gθ2), where Gθ2⊆ Λ×G3⊆ A×B ×C is the

graph of θ2and p: Λ × G3→ Λ × G3/G(3|1,2) is the natural surjection.

The fact that Γ3 and Q3 are inverse to each other is immediate from

the case n = 2. We simply apply Theorem 2.2 recursively to A × B × C =

(A × B) × C, first to G1⊆ A, G2⊆ B to obtain Λ = G{1,2}⊆ A × B, and

then to G{1,2}⊆ A × B, G3⊆ C to obtain G.

?

3.4 Remark To extend to n ≥ 4, we simply note that the 4-tuple Q2(G)

for n = 2 and 7-tuple Q3(G) for n = 3 will be replaced by a suitable (3n−2)-

tuple Qn(G), noting that 3n − 2 = 4 + 3(n − 2).

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4Application to cyclic subgroups

There are many potentially interesting applications of the (generalized)

Goursat lemma within group theory. For example, here are four immediate

applications.

4.1 Proposition Let G be a subgroup of A1× ··· × An.

(a) The subgroup G is finite if and only if G1,··· ,Gnare all finite.

(b) The subgroup G is abelian if and only if G1,··· ,Gnare all abelian.

(c) The subgroup G is p-group if and only if G1,··· ,Gnare all p-groups.

(d) The subgroup G is a sub-direct product if and only if θj= 0 for 1 ≤ j ≤

n − 1.

It is interesting to note that Hattori, in [Hat61] or [HZ09], first deter-

mined the finite subgroups of the Lie group S3(which is isomorphic to

SU(2)≈ Sp(1)≈ Spin(3) as a Lie group). He then applied Goursat’s lemma

and Proposition 4.1(a) to determine all finite subgroups of S3× S3. Using

the results of Section 3 we could now, for example, find all finite subgroups

of S3× ···

n× S3. It is also interesting that, in fact, the papers of Goursat

[Gou89] and Hattori [Hat61] or [HZ09] study closely related questions.

The next application, that of determining the cyclic subgroups of A×B,

will involve more substantial use of Goursat’s lemma . We shall henceforth

use additive notation since G1,G1,G2,G2will be abelian.

4.2 Theorem Let G be a subgroup of A × B.

(a) The subgroup G is finite cyclic if and only if G1,G2are both finite cyclic

and G1,G2have coprime order. Furthermore, |G| = lcm(|G1|,|G2|).

(b) The subgroup G is infinite cyclic if and only if either G1 is infinite

cyclic, G2 is finite cyclic, and G2 = {0}, or G2 is infinite cyclic, G1

is finite cyclic, and G1 = {0}, or both G1,G2 are infinite cyclic with

G1= G2= {0}.

Proof.

a cyclic group, must be cyclic, hence their respective subgroups G1,G2are

also cyclic.

(a) Suppose G is finite cyclic, then it is generated by an element (α,β),

whence G1 is cyclic and generated by α, G2 cyclic and generated by β.

In either case (a) or (b), G1or G2being the homomorphic image of

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Let the respective orders of G1,G2 (i.e.

gcd(m,n). Also write m = m1d, n = n1d. Then m1,n1are coprime, and

there exist x,y ∈ Z with xm1+ yn1 = 1, or equivalently xm + yn = d.

Now n(α,β) = (nα,0) ∈ G implies nα ∈ G1. Also mα = 0 ∈ G1. Hence

dα = (xm + yn)α = x(mα) + y(nα) ∈ G1. It follows that cα / ∈ G1 if

0 < c < d. For, if cα ∈ G1, then (cα,0) ∈ G. Hence (cα,0) = z(α,β) for

some integer z. Therefore zβ = 0 = (c-z)α, whence n|z and m|(c-z). Since

d divides m and n, we have d divides both z,c-z. As a result, d|c, which

is a contradiction. Thus we conclude that G1is the cyclic subgroup of G1,

generated by dα and having order m/d = m1. Similarly, G2is generated by

dβ and has order n1, so the orders of G1and G2are coprime.

Conversely, suppose |G1| = m1 is coprime to |G2| = n1, and set d =

|G1|/|G1| = |G2|/|G2|, m = m1d,n = n1d. Then d = gcd(m,n). Also

G1will be cyclic of order m, G2cyclic of order n. Using the isomorphism

θ: G1/G1→ G2/G2, choose a generator α of G1and let θ[α] = [β]. Then [α]

and [β] are both elements of of order d, whence β has order d|G2| = dn1= n

and generates G2. Also, by Goursat’s lemma, γ = (α,β) ∈ G. Finally, the

order of γ is lcm(o(α),o(β))=lcm(m,n) = (mn/d) = dm1n1. Further, again

using Goursat’s lemma, |G| = |Gθ||G1||G2| = dm1n1. Thus G is cyclic of

this order and generated by (α,β).

(b) Since G is infinite and G ⊆ G1× G2, at least one of G1,G2must be

infinite cyclic. Without loss of generality, suppose G1≈ Z. Now suppose

(α,β) generates the cyclic group G, then α generates G1, and β generates

G2. We claim that G2= {0}. For, if y ∈ G2then y = rβ for some integer

r, whence (0,y) = r(0,β) ∈ G. This implies (0,y) = k(α,β) = (kα,kβ)

for some integer k. Therefore kα = 0, whence k = 0 and y = kβ = 0.

Hence G2= {0}. We now consider separately the cases G2infinite and G2

finite (the case G1finite and G2≈ Z is symmetric to the latter, so can be

omitted).

Suppose first G ≈ Z with G2≈ Z. Then the argument in the previous

paragraph now also implies G1= {0}. Conversely, suppose G1≈ G2≈ Z

and G1= G2= {0}. Then the isomorphisms (cf. Remark 2.3)

of α,β) be m,n, and set d =

G/(G1× G2) ≈ G1/G1

θ− → G2/G2

reduce to G ≈ G1≈ G2≈ Z.

Secondly, for the remaining case, suppose G ≈ Z,G1≈ Z as before and

now G2≈ Znis cyclic of order n, n ≥ 2. Then n(α,β) = (nα,0) implies

nα ∈ G1 and clearly iα / ∈ G1 if i < n. Thus G1 ≈ nZ, and as before

G2= {0}.

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Conversely, suppose G1≈ Z,G1≈ nZ,G2≈ Znand G2= {0}. In this

case we have the isomorphism θ: G1/G1→ G2/G2≈ Zn. Let α ∈ G1with

[α] generating G1/G1. Then θ[α] = [β] = β generates G2/G2= G2≈ Zn.

We claim that G is generated by the single element (α,β), and thus is

infinite cyclic. To see this, let (x,y) ∈ G ⊆ G1× G2, so x = jα,y = kβ

for some integers j,k.Furthermore (x,y) ∈ G implies θ[x] = [y] = y,

which gives kβ = y = θ[jα] = j(θ[α]) = jβ. Then j ≡ k(mod n), so

j(α,β) = (jα,jβ) = (jα,kβ) = (x,y).

One could ask the same question for most other properties of groups, e.g.

normality, nilpotency, and solvabability to name a few. This will undoubt-

edly lead to interesting results in many cases, which are beyond the scope

of the present investigation. We conclude this note with the generalization

of Theorem 4.2 to cyclic subgroups of higher direct products.

?

4.3 Theorem Let G be a subgroup of A × B × C.

(a) The subgroup G is finite cyclic if and only if G1,G2,G3are finite cyclic

and each of the pairs of integers (|G(1|2)|,|G(2|1)|), (|G(1|3)|,|G(3|1)|),

(|G(2|3)|,|G(3|2)|) is coprime. In this case one also has

|G| = lcm(|G1|,|G2|,|G3|).

(b) The subgroup G is infinite cyclic if and only if one of the following three

cases (up to obvious permutation of indices) occur :

(i)

G1≈ Z, G2and G3are finite cyclic, G(2|1) = G(3|1) = {0},

and G(2|3),G(3|2) are coprime.

(ii)

G1 ≈ G2 ≈ Z, G3 finite cyclic, and G(2|1) = G(3|1) =

G(1|2) = G(3|2) = {0}.

(iii)Gi≈ Z for i = 1,2,3 and G(i|j) = 0 for 1 ≤ i ?= j ≤ 3.

Proof.

part (a) of Theorem 4.2 to G12gives us G1,G2finite cyclic with |G(1|2)|

coprime to |G(2|1)|. The other conditions follow by symmetry.

Conversely, suppose G1,G2,G3 are all finite cyclic with respective or-

ders m,n,p, and that the three coprimality conditions hold.

Theorem 4.2(a) two times we obtain that G12 and G12 are both finite

cyclic with respective orders lcm(m,n), lcm(|G(1|3)|,|G(2|3)|).

apply Theorem 3.3, which tells us that G is determined by the surjec-

tion θ2: G12։ G3/G(3|1,2), or equivalently by the induced isomorphism

φ: G12/G12

− → G3/G(3|1,2). A third application of Theorem 4.2(a) now

(a) If G is finite cyclic, then so is G12= π12(G) ⊆ A×B. Applying

Applying

We now

≈

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tells us that G will be cyclic if |G12| and |G(3|1,2)| are coprime.

|G12| = lcm(|G(1|3)|,|G(2|3)|), and |G(3|1,2)| is a divisor of G(3|1) which is

coprime to |G(1|3)|. Thus |G(3|1,2)| is coprime to |G(1|3)|, and similarly is

coprime to |G(2|3)|, so also coprime to their least common multiple |G12|.

(b) The three cases when G is infinite cyclic all follow from Theorem

4.2(b) in obvious ways, namely in (b)(i) we use A × B × C ≈ A × (B × C),

in (b)(ii) and (b)(iii) we use A×B×C ≈ (A×B)×C. We omit the details.

?

The generalization of this theorem to n ≥ 3 is now clear.

But

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Kristine Bauer

Department of Mathematics and Statistics

University of Calgary

Calgary, Alberta, Canada T2N 1N4

e-mail: bauerk@ucalgary.ca

Debasis Sen

Department of Mathematics and Statistics

University of Calgary

Calgary, Alberta, Canada T2N 1N4

e-mail: sen.deba@gmail.com

Peter Zvengrowski

Department of Mathematics and Statistics

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University of Calgary

Calgary, Alberta, Canada T2N 1N4

e-mail: zvengrow@ucalgary.ca

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