Goursat's Lemma Presentation

Generalizing Goursat
Notation and Constructions
Results and extentions
Bibliography and Special Thanks

Applications and Generalizations of Goursat’s
Lemma

Caridad Arroyo Sean Eggleston Boanne MacGregor

VIGRE Symposium, 2009

Arroyo, Eggleston, MacGregor Applications and Generalizations of Goursat’s Lemma


Introduction

Edouard Jean Baptiste Goursat (1858-1936) was a French
mathematician.
He described all the subgroups of the direct product of two
groups, G1 × G2 , in terms of an isomorphism between
quotient subgroups of the individual groups G1 and G2 .



Goursat’s Lemma for groups

Goursat’s Lemma
Let H be a subgroup of G1 × G2 .
2
Let H1 = {a ∈ G1 | (a, 1) ∈ H},
H1 = {a ∈ G1 | (a, b) ∈ H for some b ∈ G2 },
1
H2 = {b ∈ G2 | (1, b) ∈ H},
H2 = {b ∈ G2 | (a, b) for some a ∈ G1 ∈ H}.
Then Hij ⊆ Hi are subgroups of Gi with Hij Hi and the map
2 1 2 1
fH : H1 /H1 → H2 /H2 given by fH (aH1 ) = bH2 where (a,b) ∈ H is
an isomorphism. Moreover, if H G1 × G2 , then Hi , Hij Gi and
Hi /Hij ⊆ Z (Gi /Hij ). Moreover every subgroup of G1 × G2 arises in
this form.



Working an Example

Table: Subgroups of S3 and Z6
S3 Z6



Working an Example

S3 Z6
e e



Working an Example

S3 Z6
e e
A3 = {e, (123) , (132)} 2Z6



Working an Example

S3 Z6
e e
A3 = {e, (123) , (132)} 2Z6
C1 = {e, (1, 2)} 3Z6
C2 = {e, (1, 3)} -
C3 = {e, (2, 3)} -



Working an Example

S3 Z6
e e
A3 = {e, (123) , (132)} 2Z6
C1 = {e, (1, 2)} 3Z6
C2 = {e, (1, 3)} -
C3 = {e, (2, 3)} -
S3 Z6



Working an example

Table: Quotient Groups of S3 and Z6
S3 Z6
A3 /A3 3Z6 /3Z6
S3 /A3 3Z6 /e
A3 /e Z6 /3Z6
e/e e/e
Ci /Ci 2Z6 /2Z6
Ci /e 2Z6 /e
- Z6 /2Z6



Working an example

Table: Isomorphims between the quotient groups
Order 1 Order 2 Order 3
A3 /A3 ∼ e/e
= S3 /A3 ∼ Z6 /2Z6
= A3 /e ∼ Z6 /3Z6
=
A3 /A3 =∼ 3Z6 /3Z6 S3 /A3 ∼ 3Z6 /e
= A3 /e ∼ 2Z6 /e
=
A3 /A3 ∼ 2Z6 /2Z6
= Ci /e ∼ 3Z6 /e
= -
Ci /Ci ∼ e/e
= Ci /e ∼ e/e
= -
∼ 3Z6 /3Z6
Ci /Ci = ∼ Z6 /2Z6
Ci /e = -
Ci /Ci ∼ 2Z6 /2Z6
= - -



Working an example

We would work with the S3 /A3 ∼ Z6 /2Z6 isomorphim that
=
results from:
{{e, (123) , (132)} , {(12) , (23) , (13)}} ∼ {{0, 2, 4} , {1, 3, 5}}
=
From this product H, we obtained:
{(e, 0), (e, 2), (e, 4), ((123), 0), ((123), 2), ((123), 4), ((132), 0),
((132), 2), ((132), 4), ((12), 1), ((12), 3), ((12), 5), ((13), 1), ((13), 3),
((13), 5), ((23), 1), ((23), 3), ((23), 5)}



Working an example

∼
We will work with the isomorphim S3 /A3 = Z6 /2Z6 ;
∼ {{0, 2, 4} , {1, 3, 5}}
{{e, (123) , (132)} , {(12) , (23) , (13)}} =
From this product H, we obtained:
{(e, 0), (e, 2), (e, 4), ((123), 0), ((123), 2), ((123), 4), ((132), 0),
((132), 2), ((132), 4), ((12), 1), ((12), 3), ((12), 5), ((13), 1), ((13), 3),
((13), 5), ((23), 1), ((23), 3), ((23), 5)}



Working an example

We will try to ﬁgure out if this product yields a subgroup of
2 1
S3 × Z6 . First, notice that: H1 = A3 , H1 = S3 , H2 = e, and
H2 = Z6 . Also, A3 S3 and 2Z6 Z6 , so H S3 × Z6 .



Goals in Generalizing

To generalize to H ≤ G1 × G2 × G3 , we need to ﬁnd:
Necessary conditions
Suﬃcient conditions
Invertible construction


Denoting Subgroups
A First Attempt
Structure of a Subgroup

Notation for Three Groups

Notation generalizes to H ≤ G1 × G2 × G3 . Denote subgroups of
H:
H1 = {a ∈ G1 | (a, b, c) for some b ∈ G2 , c ∈ G3 }.
1
H2 = {b ∈ G2 | (1, b, c) ∈ H for some c ∈ G3 }.
1
H23 = {(b, c) ∈ G2 × G3 | (1, b, c) ∈ H}.
H1 = {a ∈ G1 | (a, 1, 1) ∈ H}.
Could easily generalize to G1 × G2 × · · · × Gn .


Denoting Subgroups
A First Attempt

An Easy Generalization (Not!)

Unfortunately, the obvious “Goursat’s Triangle” doesn’t work.
ˆ ¯
That is, ﬁnding all Hi Hi ≤ Gi satisfying
¯ ¯ ¯
H1 ∼ H2 ∼ H3
= = ,
ˆ
H1 ˆ
H2 ˆ
H3
does NOT yield all subgroups of G1 × G2 × G3 .

Example:
Z3 × Z5 × Z5 → 10 subgroups, 8 three-way isomorphisms.


Denoting Subgroups
A First Attempt

Recursive Goursat-ing

Applying Goursat’s Lemma to G1 × (G2 × G3 ) gives:
ϕ
H1 ∼ H23
= 1
H1 H23

We can deﬁne projections π1 : H23 → H2 and π2 : H23 → H3 .
1
H23
H
1 1
H23
H
1
2 3
The functions ϕi = πi ◦ ϕ are surjective homomorphisms:
H1
H1
e
ϕ1 ϕ2
H2 H3
1
H2 1
H3


Denoting Subgroups
A First Attempt

Recursive Goursat-ing (continued)

1 1 1
Because H23 ≤ H2 × H3 , apply Goursat’s Lemma again to ﬁnd:
1 1
H2 ∼ H3
= .
H2 H3
As in proof of Goursat’s Lemma, an isomorphism is given by

σ(b H2 ) = c H3
1
where (b, c) ∈ H23 .


Denoting Subgroups
A First Attempt

Goursat’s Recursion Pyramid

Putting this together gives a “Pyramid Lemma”:
H1
H1
e
ϕ1 ϕ2
H2 H3
1
H2 1
H3

H21 σ H31
←→
H2
e H3
e

Can permute indices to get three such diagrams.


Denoting Subgroups
A First Attempt

Isomorphisms within H

If H ≤ G1 × G2 × G3 , then
in any pair of groups Gi , Gj , there exist isomorphic quotient groups

Hi ∼ Hj
= i
Hij Hj


Denoting Subgroups
A First Attempt

More Isomorphisms within H

If H ≤ G1 × G2 × G3 , then
in the pair of G2 , G3 , there exist isomorphic quotient groups

H2 ∼ H3
3 = H2
H2 3
H2 H3 3 2
Let (a, b, c) ∈ H and let ψ : 3
H2
→ 2
H3
be given by ψ(bH2 ) = cH3 .

Then ψ is an isomorphism. (Again, similar to proof of Goursat’s
Lemma.)


Denoting Subgroups
A First Attempt

Even More Isomorphisms within H

All three pairs together give a “disjoint triangle”:
H1 H1
2
H1 3
H1

H2 H3
1
H2 1
H3
H2 H3
3
H2
−→ 2
H3


Denoting Subgroups
A First Attempt

Isomorphisms and Homomorphisms within H

Connecting a triangle side and the pyramid bottom gives the
commutative diagram:
H21 σ H31
−→
H2
e H3
e
↓ ρ1 ↓ ρ2
H2 ψ H3
3
H2
−→ 2
H3

σ, ψ are isomorphisms, ρi are homomorphisms.
If (g1 , g2 , g3 ) ∈ H, gi ∈ Hi1 , then ρi (gi Hi ) = gj Hji (i, j = 2, 3).
σ ◦ ρ2 = ρ1 ◦ ψ 2


Denoting Subgroups
A First Attempt

The Next Step

What do we do with all this???


Notation and Constructions Conjectural History
Results and extentions Board Time for Computational Examples

Conjectural History

1. Recursive Goursat
2. The obvious ﬁrst attempt: “Goursat’s Triangle”
3. Final



Possible Cases for Generalization

All Groups
Finite Groups
Finite Cyclic Groups



Necessary, not Suﬃcient

H1 H1
2
H1 3
H1

H2 H3
1
H2 1
H3
H2 H3
3
H2
−→ 2
H3



Board Time for Computational Examples

Consider: Z2 xZ2 xZ2

For the subgroup generated by: (1, 1, 0), (1, 0, 1), (0, 1, 1)

Is the Disjoint Triangle suﬃcient?



Summation of Hope

12 subgroups (Hi , Hij , Hi )
˜
Isomorphic quotient groups
Constructed homomorphisms


Notation and Constructions Bibliography
Results and extentions Special Thanks

Anderson, Dan D., and Vic Camillo. “Subgroups of Direct
Products of Groups, Ideals and Subrings of Direct Products of
Rings, and Goursat’s Lemma.” Contemporary Mathematics
480 (2009): 1-11. Print.


Notation and Constructions Bibliography
Results and extentions Special Thanks

Acknowledgements

Many thanks to our mentors,

Dr. Dan Anderson,
Dr. Paul-Hermann Zieschang
and
Mr. Carlos De la Mora

for their tireless work and dedication.


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