This document summarizes research attempting to generalize Goursat's Lemma, which describes subgroups of direct products, to n groups. The researchers analyzed the case of 3 groups and defined subsets of each group involved in the direct product. They proved certain projections formed a group but found counterexamples preventing a full generalization. Diagrams of isomorphisms between quotient groups were explored but did not satisfy all conditions. In conclusion, directly generalizing the isomorphism or using disjoint triangles was insufficient, and fully generalizing the cosset representations between groups remained an open problem.

1. Generalizations and Applications of Goursat’s Lemma
The University of Iowa
Mathematics Department
Dr. Dan Anderson
Caridad Arroyo Quijano
A bstract
Goursat’s Lemma describes subgroups of a direct product in terms of normal
subgroups of the individual groups. This statement also gives an isomorphism of the form
f: H12/H11→ H22/H21 where G1 and G2 are groups, Hi2 ≤ Gi, and Hi1  Hi2. We would like
to generalize Goursat’s Lemma to find the subgroups of G1 × G2 × G3 × . . . × Gn. Of
course, we could do this by applying the lemma recursively to G1 × (G2 × G3 × . . . ×
Gn); however, this quickly grows convoluted and turns out an inelegant solution. We
instead wish to generate the subgroups of G1 × . . . × Gn in terms of the subgroups of G1,
G2, . . . , Gn independently. Specifically, we examine the case where n = 3.
I. I ntroduction
During you mathematical development you would definitely find the ordered
Cartesian pair product in many applications. The operation among these pairs is known as
the Cross Product ( x, y ) × ( p, q ) = ( xp, yq ) . Some applications need to compute all the
ordered pairs given two sets or groups. But, in order to simplify its calculations; how is
this binary operation algebra? Now, this becomes an abstract algebra problem.
Our goal was to generalize this computation problem for n groups, G1 × G2 × G3
× . . . × Gn. Given the statement:
Let G1 and G2 be groups and let H be a subgroup of G1 × G2 such that the
two projections ρ1 : H → G1 and ρ 2 : H → G2 are surjective. Let N ' be the kernel
of ρ1 and N the kernel of ρ 2 . One can identify N as a normal subgroup of G1 and N '
as a normal subgroup of G2 . Then the image of H is G1 N × G2 N ' is the graph of
an isomorphism G1 N ≈ G2 N ' .

2. In order to solve this problem we made reference to Edouard Jean Baptiste
Goursat, a Frech Mathematician, who found an algorithm that solved this for two groups
case, G1 and G2. This Lemma is known as Goursat’s Lemma for Groups.
Lemma 1: Goursat’s Lemma for Groups
Let H be a subgroup of G1 × G2 .
Let H12 = {a ∈ G1 | (a, 1 ) ∈ H},
H1 = {a ∈ G1 | (a, b) ∈ H for some b ∈ G2 },
H 2 = {b ∈ G2 | ( 1, b) ∈ H},
1
H 2 = {b ∈ G2 | (a, b) for some a ∈ G1 ∈ H}.
Then H i j ⊆ Hi are subgroups of Gi with H i j  H i and the map
f H : H1 /H12 → H 2 /H 2 given by f H (aH12 ) = bH 2 where (a,b) ∈ H is
1 1
an isomorphism. Moreover, if H  G1 × G2 , then H i ,H j  Gi and
H i /H i j ⊆ Z(Gi /H i j ) . Moreover every subgroup of G1 × G2 arises in
this form.
In order to generalize we first considered to review the case for n= 3 and the
Lemma restrictions: the creation of every sub groups, and invertible functions. To
simplify the sufficient condition we defined these sets, where 1 is the identity of the
binary operation.
H1 = {a ∈ G1 | (a, b, c) for some b ∈ G 2 , c ∈ G 3 }.
H 2 = {b ∈ G 2 | ( 1, b, c) ∈ H for some c ∈ G3 }.
1
H 23 = {(b, c) ∈ G2 × G3 | ( 1, b, c) ∈ H}.
1
~
H 1 = {a ∈ G1 | (a, 1, 1 ) ∈ H}.
Where we could easily generalized for G1 × G2 × · · · × Gn .
~ ~ ~ 1
Where H 1  H1, H 2  H 2  H 2 , H 3  H 3  H 3 , H 23  H 23
1 1
II. M ethodology

3. Our first though trying to solve this problem was to directly apply the Lemma
conjecture creating isomorphism between the quotient groups of the three groups. This
− − −
H1 H3
≅ H2
− ^
≅
would be for H i  H i ≤ Gi satisfying ^
H1
^
H2
^
H 3 , but this doesn’t work
because you won’t get all the subgroups. For this we found a counter example in
Ζ 3 × Ζ 5 × Ζ 5 ≈ Ζ15 × Ζ 5 where we have 8 and 10 subgroups respectively.
The next step consisted in the possibility using Goursat’s Lemma in a recursively
way, G1 × ( G2 × G3 ) ; not considered as an elegant solution since is still being complex. So
knowing the possibility of the recursion we worked breaking the product between
G2 and G3 , the isomorphism ϕ : H H ≅ H H 1 using homomorphic projections
~
1 1 23 23
ϕ1 and ϕ 2 to G2 and G3 respectively where π 1 : H 23 H 23 → H 2 H 2 and
1 1
π 2 : H 23 H 23 → H 3 H 3 are surjective and ϕ i :π i  ϕ . Why only homomorphism and not
1 1
~
isomorphism? Because H1 and H 1 are our independent variables and by function
definition you will have every element from the first group. The isomorphism between
the two others groups is necessary in order to obtain every element from both groups and
still having an inverse if you try to go back though the projections. Next, is our proof
demonstrating that we only need homomorphism between every group and you
independent group.
Statement: Is the set
 ~  1  ~ 
K = { ( x, y, z ) ∈ H1 × H 2 × H 3 | ϕ1  x H 1  = y H 2 and ϕ1  x H 1  = zH 3
1
( ) } a group?
   
Proof: Let ( a1 , b1 , c ) ∈ K also let be ( a 2 , b2 , c 2 ) ∈ K we want to proof if
(a a
1
−1
2 ) (
, b1b2 1 , c1c 2 1 ∈ K . ϕ1 a1a2 1 H1
− − −
) ( −
)
= ϕ1 a1 H11a2 1 H1 = ϕ1 ( a1 H1 ) ϕ1 a2 1 H1
−
( )
= ϕ1 ( a1 H1 )ϕ1 ( a2 H 2 ) = b1 H 2 ( b2 H 2 )
−1 −1
= b1b2 H 2 .

4. ϕ 2 ( a1a2 1 H1 ) = ϕ 2 ( a1 H1a2 1 H1 ) = ϕ 2 ( a1 H1 )ϕ 2 ( a2 1 H1 ) = ϕ 2 ( a1 H1 )ϕ 2 ( a2 H1 )
− − − −1
= c1 H 3 ( c2 H 3 ) = c1c2 1 H 3
−1 −
Hence, K is a group demonstrating the homomorphism of these projections. From this
conjecture and our created sets we have the diagram called Recursive Goursat -ing:
Figure 1: Recursive – Goursat-ing
As given in our statement we can easily see that there’s no isomorphism between
1 1
H 2 H 2 and H 3 H 3 . Applying Goursat’s Lemma again in this broken recursively we
can see that for H 23 ≤ H 2 × H 3 so our isomorphism would be between
1 1 1
~ ~
H 2 H 2 ≅ H 3 H 3 using the Correspondence Theorem obtaining the Pyramid Lemma
1 1
diagram.
Figure 2: Pyramid Lemma
Because we focused our project in finite cyclic groups, trying to make a
generalization we checked the different cases for the generators. For this we check the
three cases: o( x0 ) = o( y0 ) = o( zo ) , o( xo ) = o( yo ) ≠ o( zo ) and o( x0 ) ≠ o( y0 ) ≠ o( zo ) .
For H = ( xo , yo , zo ) ⊆ H1 × H 2 ×H 3
∫ Case 1: Let o( xo ) = o( yo ) = o( zo ) and y∈ H1 such that (1, y,z ) ∈ H ⇒ x o , y o ,z o
2
k k k
( )

5. ~ ~
= (1, y,z ) = (1,1,1) ⇒ H 2 = H 2 = {1} . Where a proof for H 3 = H 3 is analogous. Hence, if
o( x0 ) = o( y0 ) = o( zo ) , then H 2 = H 1 = { e} and H 3 = H 1 = { e} .
~ ~
2 3
∫ Case 2: Let o( xo ) = o( yo ) ≠ o( zo ) and z∈ H 3 we get that there exists y such that
(1, y, z,) ∈ H , but since H is cyclic
(1, y, z ) = ( xo , y o , z o ) = xo = y o = 1 ⇒ (1,1, z o )∈ H ⇒ (1,1, z ) ∈ H where
k k k k k k ~ ~
1
z∈H 3 ⇒ H 3 = H3 = 1 .
( ) ~
( )
Let y ' ∈ H 1 then there exists z ' ∈ H 3 such that 1, y; , z ' ∈ H , but H 3 = H 3 ⇒ 1, y ' ,1 ∈ H .
2
1 1
Hence, for o( xo ) = o( yo ) ≠ o( zo ) then, H 2 = H 2 = { e} and H 3 = H 3 = { e} .
~ ~
1 1
~
∫ Case 3: We founded a counter example in Ζ15 × Z10 × Ζ6 where our H 2 = { 0} and
~
H 1 = { 0,5} and H = { 0} and H 3 = { 0,3} . This last conclusion complicated our work,
1
2 3
since we can’t make the generalization simplifying our quotient groups in order to
create the isomorphism to satisfy the completeness condition.
To continue with the generalization we tried to use a theorem in order to use
1 1
cosets representations to create the isomorphism between H 2 H 2 and H 3 H 3 and satisfy
~
the completeness condition having every element, but H i as a normal subgroup.
If you apply the Lemma again you can see that for H ≤ G1 × G2 × G3 you could
find for any pair of groups Gi and G j there exist isomorphic quotient groups
H i H i j ≅ H j H ij creating a series of isomorphism defined by ψ ( bH i j ) = cH ij
for ( a, b, c ) ∈ H . Representing all the possibilities we created the next diagram on Figure
3. This diagram doesn’t satisfy the completeness condition. You can easily see this with
the counter example Ζ 2 × Ζ 2 × Ζ 2 with (1,1,1) as a generator.

6. Figure 3: Disjoint Triangle
Concentrated creating in creating isomorphism if you make a partial union
between the Figure 3 and the isomorphism in Figure 2 you’ll get the commutative
diagram on Figure 4.
Figure 4: Isomorphism and Homomorphism
Here we have that σ and ψ are isomorphism and ρ1 and ρ 2 are homomorphism
 ~

where for ( g1 , g 2 , g 3 ) ∈ H , and g i ∈ H i ⇒ ρ i  g i H i  = g j H j . Also we have that
1 i
 
( ) ( )  
( ) 
( 
)
~ ~
ρ1 ψ = σ  ρ 2 . ⇒ ρ1  g 2 H 2 = σ  g 3 H 32 ⇒  g 2 H 2  g 3 H 32 =  g 2 H 2  g 3 H 32 . An easy
3
   
σ
way to visualize this would be (1, b,1)  (1, b, c ) → (1,1, c )  (1, b, c )
ρ1 ↓ ↓ ρ2
ψ
( b,1)  ( b, c ) → (1, c )  ( b, c )
III. Conclusion
In conclusion we worked in sets creation, defining functions and proving if they
were well defined trying to satisfy the Lemma conditions. We couldn’t make a
1 1
generalization for the isomorphism between H 2 H 2 and H 3 H 3 because we didn’t
succeed creating the conjecture between the cossets representative for G2 and G3 which
it seems to be possible with more time and effort. We found that the direct conjecture

7. creating isomorphism between the three groups doesn’t work and a series of isomorphism
between the different quotient groups which aren’t sufficient because it doesn’t satisfy
the completeness condition.