lattice of subgroups

lattice of subgroups

Let

G

be a group and

L

⁢

(

G

)

be the set of all

subgroups

of

G

. Elements of

L

⁢

(

G

)

can be ordered by the

set inclusion

relation

⊆

. This way

L

⁢

(

G

)

becomes a

partially ordered set

.

For any

H

,

K

∈

L

⁢

(

G

)

, define

H

∧

K

by

H

∩

K

. Then

H

∧

K

is a subgroup of

G

and hence an element of

L

⁢

(

G

)

. It is not hard to see that

H

∧

K

is the largest subgroup of both

H

and

K

.

Next, let

X

=

H

∪

K

and define

H

∨

K

by

⟨

X

⟩

, the subgroup of

G

generated by

X

. So

H

∨

K

∈

L

⁢

(

G

)

. Each

element

in

H

∨

K

is a finite

product

of elements from

H

and

K

. Again, it is easy to see that

H

∨

K

is the smallest subgroup of

G

that has

H

and

K

as its subgroups.

With the two

binary operations

∧

and

∨

,

L

⁢

(

G

)

becomes a

lattice

. It is a

bounded lattice

, with

G

as the top element and

⟨

e

⟩

as the bottom element. Furthermore, if

{

H

i

∣

i

∈

I

}

is a set of subgroups of

G

indexed by

some set

I

, then both

⋀

i

∈

I

H

i

and

⋁

i

∈

I

H

i

are subgroups of

G

. So

L

⁢

(

G

)

is a

complete lattice

. From this, it is easy to produce a lattice which is not a

subgroup lattice

of any group.

Atoms in

L

⁢

(

G

)

, if they exist, are finite

cyclic groups

of prime order (or

ℤ

/

p

⁢

ℤ

, where

p

is a prime), since they have no non-trivial

proper subgroups

.

Remark.

Finding lattices of subgroups of groups is one way to classify groups. One of the main results in this branch of

group theory

states that the lattice of subgroups of a group

G

is

distributive

(

http://planetmath.org/DistributiveLattice

) iff

G

is

locally cyclic

.

It is generally not true that the lattice of subgroups of a group determines the group up to

isomorphism

. Already for groups of order

p

3

,

p

>

2

or

p

4

, for all primes, there are

examples of groups

with isomorphic (

http://planetmath.org/LatticeIsomorphism

) subgroup lattices which are not

isomorphic groups

.

Example.

Note that

Aut

⁡

ℤ

p

2

≅

ℤ

p

-

1

×

ℤ

p

. Therefore it is possible to from a non-trivial

semidirect product

ℤ

p

2

⋊

ℤ

p

. The lattice of subgroups of

ℤ

p

2

⋊

ℤ

p

is the same as the lattice of subgroups of

ℤ

p

2

×

ℤ

p

. However,

ℤ

p

2

⋊

ℤ

p

is

non-abelian

while

ℤ

p

2

×

ℤ

p

is

abelian

so the two groups are not isomorphic.

Similarly, the groups

ℤ

p

i

⋊

ℤ

p

and

ℤ

p

i

×

ℤ

p

for any

i

>

2

and any primes

p

also have isomorphic subgroup lattices while one is non-abelian and the other abelian. So this is indeed a family of counterexamples.

Upon inspecting these example it becomes clear that the

non-abelian groups

have a different sublattice of

normal subgroups

. So the question can be asked whether two

groups with

isomorphic subgroup lattices including matching up

conjugacy classes

(so even stronger than matching normal subgroups) can be non-isomorphic groups. Surprisingly the answer is yes and was the dissertation of Ada Rottländer

[

1

]

, a student of Schur’s, in 1927. Her example uses groups already discovered by Otto Hölder in his famous classification of the groups of order

p

3

,

p

2

⁢

q

, and

p

4

. With the modern understanding of groups the counterexample is rather simple to describe – though a proof remains a little tedious.

Let

V

=

ℤ

q

2

where

q

is a prime – that is

V

is the 2-dimensional

vector space

over the field

ℤ

q

. Let

p

|

q

-

1

be another prime. As

p

|

q

-

1

,

ℤ

p

≤

ℤ

q

×

so if we write

ℤ

p

=

⟨

ω

⟩

multiplicatively so that we have for every

n

∈

ℤ

q

,

n

↦

ω

⋅

n

is an

automorphism

of

ℤ

q

.
(Note that

ω

is often called a

primitive

p

-th root of unity in

ℤ

q

as it spans the

ℤ

p

subgroup of

ℤ

q

×

.) Furthermore, for any

0

≤

i

≤

p

-

1

we get an automorphism

f

i

:

ℤ

q

→

ℤ

q

given by

f

i

⁢

(

n

)

=

ω

i

⁢

n

.

Therefore to every

0

≤

i

≤

p

-

1

we can define a group

G

i

=

⟨

V

,

g

i

⟩

,

g

i

=

f

1

⊕

f

i

as a subgroup of

A

⁢

G

⁢

L

⁢

(

V

)

. That is to say,

G

i

=

V

⋊

⟨

g

i

⟩

where the action of

g

i

on

V

is given by: for every

v

∈

V

,

v

=

[

n

m

]

for

n

,

m

∈

ℤ

q

set

g

i

⁢

(

v

)

=

[

ω

0

0

ω

i

]

⁢

[

n

m

]

=

[

ω

⁢

n

ω

i

⁢

m

]

.

We are now prepared to give the Rottländer counterexample.

Now let

a

and

b

be integers between

2

and

p

-

1

such that

a

is not

congruent

to

b

modulo

p

. Notice this already forces

p

>

3

so our smallest example is

q

=

11

and

p

=

5

. Then

G

a

is not isomorphic to

G

b

(compare the

eigenvalues

of

g

a

to

g

b

– they are not equal so the

linear transformations

are not

conjugate

in

G

⁢

L

⁢

(

2

,

q

)

.) However,

G

a

and

G

b

have isomorphic subgroup lattices including matching conjugacy classes.

References

1

Rottländer, Ada,

Nachweis der Existenz nicht-isomorpher Gruppen von gleicher
Situation der Untergruppen

,
Math. Z. vol. 28, 1928, 1, pp. 641– 653, ISSN 0025-5874.
MR MR1544982,

Title

lattice of subgroups

Canonical name

LatticeOfSubgroups

Date of creation

2013-03-22 15:47:42

Last modified on

2013-03-22 15:47:42

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

14

Author

CWoo (3771)

Entry type

Definition

Classification

msc 20E15

Synonym

subgroup lattice

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