aarjaneiro/planetmath huggingface dataset:
Erdős-Ginzburg-Ziv theorem

Erdős-Ginzburg-Ziv theorem

If

a

1

,

a

2

,

…

,

a

2

⁢

n

-

1

is a set of integers, then there exists a subset

a

i

1

,

a

i

2

,

…

,

a

i

n

of

n

integers

such that

a

i

1

+

a

i

2

+

⋯

+

a

i

n

≡

0

(

mod

n

)

.

The theorem is also known as the

EGZ theorem

.

References

1

Melvyn B. Nathanson.

Additive Number Theory

: Inverse Problems and Geometry of
Sumsets

, volume 165 of

GTM

.

Springer, 1996.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0859.11003

Zbl 0859.11003.

2

Hao,P.

On a

Congruence

modulo a Prime

Amer. Math. Monthly, vol. 113, (2006), 652-654

Title

Erdős-Ginzburg-Ziv theorem

Canonical name

ErdHosGinzburgZivTheorem

Date of creation

2013-03-22 13:40:00

Last modified on

2013-03-22 13:40:00

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

7

Author

bbukh (348)

Entry type

Theorem

Classification

msc 20D60

Classification

msc 11B50

Synonym

EGZ theorem

Generated on Fri Feb 9 19:24:37 2018 by

LaTeXML
non-commuting graph

non-commuting graph

Let

G

be a

non-abelian group

with center

Z

⁢

(

G

)

. Associate a graph

Γ

G

with

G

whose vertices are the non-central

elements

G

∖

Z

⁢

(

G

)

and whose edges join those vertices

x

,

y

∈

G

∖

Z

⁢

(

G

)

for which

x

⁢

y

≠

y

⁢

x

. Then

Γ

G

is said to be
the

non-commuting graph

of

G

.
The non-commuting graph

Γ

G

was first considered by Paul
Erdös, when he posed the following problem in 1975 [B.H. Neumann,

A problem of Paul Erdös on
groups,

J. Austral. Math. Soc. Ser. A

21

(1976), 467-472]:

Let

G

be a group whose non-commuting graph has no

infinite

complete

subgraph

. Is it true that there is a finite bound on the

cardinalities

of complete subgraphs of

Γ

G

?

B. H. Neumann answered positively Erdös’ question.

The non-commuting graph

Γ

G

of a non-abelian group

G

is always

connected

with

diameter

2 and girth 3.
It is also

Hamiltonian

.

Γ

G

is planar if and only if

G

is isomphic to the

symmetric group

S

3

, or
the

dihedral group

𝒟

8

of order

8

or the

quaternion group

Q

8

of order

8

.

See
[Alireza Abdollahi, S. Akbari and H.R. Maimani,

Non-commuting graph of a group

, Journal of Algbera,

298

(2006) 468-492.] for proofs of these properties of

Γ

G

.

Examples

Symmetric group

S

3

The symmetric group

S

3

is the smallest non-abelian group. In

cycle
notation

, we have

S

3

=

{

(

)

,

(

12

)

,

(

13

)

,

(

23

)

,

(

123

)

,

(

132

)

}

.

The center is

trivial

:

Z

⁢

(

S

3

)

=

{

(

)

}

. The non-commuting graph in
Figure

1

contains all possible edges except one.

Figure 1:

Non-commuting graph of the symmetric group

S

3

Octic group

The dihedral group

𝒟

8

, generally known as the

octic group

, is
the

symmetry group

of the

http://planetmath.org/node/1086

square. If you label the vertices of the
square from

1

to

4

going along the edges, the octic group may be
seen as a

http://planetmath.org/node/1045

subgroup

of the symmetric group

S

4

:

𝒟

8

:=

{

(

)

,

(

13

)

,

(

24

)

,

(

12

)

⁢

(

34

)

,

(

13

)

⁢

(

24

)

,

(

14

)

⁢

(

23

)

,

(

1234

)

,

(

1432

)

}

.

So the octic group has

http://planetmath.org/node/2871

order

8

(hence its name), and its center
consists of the

identity element

and the simultaneous flip around both
diagonals

(

13

)

⁢

(

24

)

. Its non-commuting graph is given in
Figure

2

.

Figure 2:

Non-commuting graph of the octic group

Title

non-commuting graph

Canonical name

NoncommutingGraph

Date of creation

2013-03-22 15:18:50

Last modified on

2013-03-22 15:18:50

Owner

GrafZahl (9234)

Last modified by

GrafZahl (9234)

Numerical id

11

Author

GrafZahl (9234)

Entry type

Definition

Classification

msc 20D60

Classification

msc 05C25

Synonym

non-commuting graph of a group

Related topic

NonAbelianTheories

Related topic

NonAbelianStructures

Related topic

CommutativeVsNonCommutativeDynamicModelingDiagrams

Related topic

GeneralizedToposesTopoiWithManyValuedLogicSubobjectClassifiers

Generated on Fri Feb 9 19:24:40 2018 by

LaTeXML
Order Conjecture for non-commuting graph of a group

Order Conjecture for non-commuting graph of a group

The following was conjectured by A. Abdollahi, S. Akbari and H. R. Maimani
in (

Non-commuting graph

of a group, Journal of Algebra, 298 (2006) 468-492.)

Order

Conjecture

.

If

G

and

H

are two

non-abelian

finite groups

with

isomorphic

non-commuting graphs, then

|

G

|

=

|

H

|

.

It was proved that Order Conjecture is true if and only if it is true for all non-abelian

solvable

finite

A

⁢

C

-groups.
By an

A

⁢

C

-group, we mean a group in which the

centralizer

of every non-central element is

abelian

.

The order Conjecture has been refuted in the following paper

[*] A. R. Moghaddamfar, On non-commutating graphs, Siberian Math. J.

47

(2006), no. 5, 911-914.

It is mentioned in [*] that the example given in the article is due to M. Isaacs.

Title

Order Conjecture for non-commuting graph of a group

Canonical

name

OrderConjectureForNoncommutingGraphOfAGroup

Date of creation

2013-03-22 15:18:53

Last modified on

2013-03-22 15:18:53

Owner

abdollahi (9611)

Last modified by

abdollahi (9611)

Numerical id

10

Author

abdollahi (9611)

Entry type

Conjecture

Classification

msc 20D60

Generated on Fri Feb 9 19:24:42 2018 by

LaTeXML
Cauchy’s theorem

Cauchy’s theorem

Let

G

be a

finite group

and let

p

be a prime dividing

|

G

|

. Then there is an element of

G

of order

p

.

Title

Cauchy’s theorem

Canonical name

CauchysTheorem

Date of creation

2013-03-22 12:13:05

Last modified on

2013-03-22 12:13:05

Owner

Evandar (27)

Last modified by

Evandar (27)

Numerical id

5

Author

Evandar (27)

Entry type

Theorem

Classification

msc 20D99

Classification

msc 20E07

Generated on Fri Feb 9 19:24:47 2018 by

LaTeXML
estimation of index of intersection subgroup

estimation of index of intersection subgroup

Theorem

.

If

H

1

,

H

2

,

…

,

H

n

are

subgroups

of

G

, then

[

G

:

⋂

i

=

1

n

H

i

]

≦

∏

i

=

1

n

[

G

:

H

i

]

.

Proof.

We prove here only the case

n

=

2

;  the general case may be handled by the

induction

.

Let

H

1

∩

H

2

:=

K

.  Let

R

be the set of the

right cosets

of

K

and

R

i

the set of the right cosets of

H

i

(

i

=

1

,

2

).  Define the

relation

ϱ

from

R

to

R

1

×

R

2

as

ϱ

:=

{

(

K

x

,

(

H

1

x

,

H

2

x

)

)

⋮

x

∈

G

}

.

We then have the

equivalent

(

http://planetmath.org/Equivalent3

) conditions

K

⁢

x

=

K

⁢

y

,

x

⁢

y

-

1

∈

K

,

x

⁢

y

-

1

∈

H

1

∧

x

⁢

y

-

1

∈

H

2

,

H

1

⁢

x

=

H

1

⁢

y

∧

H

2

⁢

x

=

H

2

⁢

y

,

(

H

1

⁢

x

,

H

2

⁢

x

)

=

(

H

1

⁢

y

,

H

2

⁢

y

)

,

whence

ϱ

is a

mapping

and

injective

,

ϱ

:

R

→

R

1

×

R

2

.  i.e. it is a

bijection

from

R

onto the subset

{

ϱ

⁢

(

x

)

⁢

⋮

⁢

x

∈

R

}

of

R

1

×

R

2

.  Therefore,

card

⁡

(

R

)

≦

card

⁡

(

R

1

×

R

2

)

=

card

⁡

(

R

1

)

⋅

card

⁡

(

R

2

)

.

As a

consequence

one obtains the

Theorem (Poincaré

).  The index of the

intersection

of finitely many subgroups with finite indices (

http://planetmath.org/Coset

) is finite.

Title

estimation of index of intersection subgroup

Canonical name

EstimationOfIndexOfIntersectionSubgroup

Date of creation

2013-03-22 18:56:46

Last modified on

2013-03-22 18:56:46

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 20D99

Synonym

index of intersection subgroup

Related topic

LogicalAnd

Related topic

Cardinality

Generated on Fri Feb 9 19:24:50 2018 by

LaTeXML
example of fully invariant subgroup

example of fully invariant subgroup

The

derived subgroup

[

G

,

G

]

is a

fully invariant subgroup

because if

f

is an

endomorphism

of

G

, then for each word of

commutators

[

a

1

,

b

1

]

⁢

[

a

2

,

b

2

]

⁢

⋯

⁢

[

a

m

,

b

m

]

, we have

f

⁢

(

[

a

1

,

b

1

]

⁢

[

a

2

,

b

2

]

⁢

⋯

⁢

[

a

m

,

b

m

]

)

=

[

f

⁢

a

1

,

f

⁢

b

1

]

⁢

[

f

⁢

a

2

,

f

⁢

b

2

]

⁢

⋯

⁢

[

f

⁢

a

m

,

f

⁢

b

m

]

∈

[

G

,

G

]

i.e. the

homomorphic image

of a word of commutators is a word of commutators.

Title

example of fully invariant subgroup

Canonical name

ExampleOfFullyInvariantSubgroup

Date of creation

2013-03-22 16:04:41

Last modified on

2013-03-22 16:04:41

Owner

juanman (12619)

Last modified by

juanman (12619)

Numerical id

5

Author

juanman (12619)

Entry type

Example

Classification

msc 20D99

Related topic

FullyInvariantSubgroup

Generated on Fri Feb 9 19:24:54 2018 by

LaTeXML
fully invariant subgroup

fully invariant subgroup

A

subgroup

H

of a group

G

is

fully invariant

if

f

⁢

(

H

)

⊆

H

for all

endomorphisms

f

:

G

→

G

. Such a subgroup is also called

fully characteristic

.

This is a

stronger

condition than being a

characteristic subgroup

.

The

derived subgroup

is fully invariant.

Title

fully invariant subgroup

Canonical name

FullyInvariantSubgroup

Date of creation

2013-03-22 13:13:07

Last modified on

2013-03-22 13:13:07

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

6

Author

mclase (549)

Entry type

Definition

Classification

msc 20D99

Related topic

CharacteristicSubgroup

Related topic

SubnormalSubgroup

Related topic

ExampleOfFullyInvariantSubgroup

Defines

fully invariant

Defines

fully characteristic

Defines

fully characteristic subgroup

Generated on Fri Feb 9 19:24:56 2018 by

LaTeXML
Lagrange’s theorem

Lagrange’s theorem

Let

G

be a

finite group

and let

H

be a

subgroup

of

G

. Then the order of

H

divides the order of

G

.

Title

Lagrange’s theorem

Canonical name

LagrangesTheorem

Date of creation

2013-03-22 12:13:03

Last modified on

2013-03-22 12:13:03

Owner

Evandar (27)

Last modified by

Evandar (27)

Numerical id

6

Author

Evandar (27)

Entry type

Theorem

Classification

msc 20D99

Related topic

Group

Related topic

ProofOfFermatsLittleTheoremUsingLagrangesTheorem

Related topic

ProofOfEulerFermatTheoremUsingLagrangesTheorem

Generated on Fri Feb 9 19:24:58 2018 by

LaTeXML
proof of Cauchy’s theorem in abelian case

proof of Cauchy’s theorem in abelian case

Suppose

G

is

abelian

and the order of

G

is

h

. Let

g

1

,

g

2

,

…

,

g

h

be the elements of

G

, and for

i

=

1

,

…

,

h

, let

a

i

be the order of

g

i

.

Consider the

direct sum

H

=

⊕

i

=

1

h

ℤ

/

a

i

⁢

ℤ

.

The order of

H

is obviously

a

1

⁢

a

2

⁢

⋯

⁢

a

h

. We can define a

group homomorphism

θ

from

H

to

G

by

(

x

1

,

…

,

x

h

)

↦

g

1

x

1

⁢

⋯

⁢

g

h

x

h

.

θ

is certainly

surjective

. So

|

H

|

=

|

G

|

⋅

|

ker

⁡

(

θ

)

|

. Since

p

is a

prime factor

of

G

,

p

divides —H—, and therefore must divide one of the

a

i

’s, say

a

1

. Then

g

1

a

1

/

p

is an element of order

p

.

Title

proof of Cauchy’s

theorem

in abelian case

Canonical name

ProofOfCauchysTheoremInAbelianCase

Date of creation

2013-03-22 14:30:28

Last modified on

2013-03-22 14:30:28

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

7

Author

kshum (5987)

Entry type

Proof

Classification

msc 20D99

Classification

msc 20E07

Generated on Fri Feb 9 19:25:00 2018 by

LaTeXML
proof of Lagrange’s theorem

proof of Lagrange’s theorem

We know that the cosets

H

⁢

g

form a

partition

of

G

(see the coset entry for proof of this.) Since

G

is finite, we know it can be completely decomposed into a finite number of cosets. Call this number

n

. We denote the

i

th coset by

H

⁢

a

i

and write

G

as

G

=

H

⁢

a

1

∪

H

⁢

a

2

∪

⋯

∪

H

⁢

a

n

since each coset has

|

H

|

elements, we have

|

G

|

=

|

H

|

⋅

n

and so

|

H

|

divides

|

G

|

, which proves Lagrange’s theorem.

□

Title

proof of Lagrange’s theorem

Canonical name

ProofOfLagrangesTheorem

Date of creation

2013-03-22 12:15:47

Last modified on

2013-03-22 12:15:47

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

6

Author

akrowne (2)

Entry type

Proof

Classification

msc 20D99

Generated on Fri Feb 9 19:25:02 2018 by

LaTeXML
proof of the converse of Lagrange’s theorem for finite cyclic groups

proof of the converse of Lagrange’s theorem for finite cyclic groups

The following is a proof that, if

G

is a finite

cyclic group

and

n

is a

nonnegative integer

that is a

divisor

of

|

G

|

, then

G

has a

subgroup

of order

n

.

Proof.

Let

g

be a

generator

of

G

. Then

|

g

|

=

|

⟨

g

⟩

|

=

|

G

|

. Let

z

∈

ℤ

such that

n

⁢

z

=

|

G

|

=

|

g

|

. Consider

⟨

g

z

⟩

. Since

g

∈

G

, then

g

z

∈

G

. Thus,

⟨

g

z

⟩

≤

G

. Since

|

⟨

g

z

⟩

|

=

|

g

z

|

=

|

g

|

gcd

⁡

(

z

,

|

g

|

)

=

n

⁢

z

gcd

⁡

(

z

,

n

⁢

z

)

=

n

⁢

z

z

=

n

, it follows that

⟨

g

z

⟩

is a subgroup of

G

of order

n

.
∎

Title

proof of the converse of Lagrange’s theorem for finite cyclic groups

Canonical name

ProofOfTheConverseOfLagrangesTheoremForFiniteCyclicGroups

Date of creation

2013-03-22 13:30:27

Last modified on

2013-03-22 13:30:27

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20D99

Related topic

CyclicRing3

Related topic

ProofThatGInGImpliesThatLangleGRangleLeG

Generated on Fri Feb 9 19:25:05 2018 by

LaTeXML
proof that every group of prime order is cyclic

proof that every group of prime order is cyclic

The following is a

proof that every group of prime order is cyclic

.

Let

p

be a prime and

G

be a group such that

|

G

|

=

p

. Then

G

contains more than one element. Let

g

∈

G

such that

g

≠

e

G

. Then

⟨

g

⟩

contains more than one element. Since

⟨

g

⟩

≤

G

, by Lagrange’s theorem,

|

⟨

g

⟩

|

divides

p

. Since

|

⟨

g

⟩

|

>

1

and

|

⟨

g

⟩

|

divides a prime,

|

⟨

g

⟩

|

=

p

=

|

G

|

. Hence,

⟨

g

⟩

=

G

. It follows that

G

is cyclic.

Title

proof that every group of prime order is cyclic

Canonical name

ProofThatEveryGroupOfPrimeOrderIsCyclic

Date of creation

2013-03-22 13:30:55

Last modified on

2013-03-22 13:30:55

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20D99

Related topic

ProofThatGInGImpliesThatLangleGRangleLeG

Generated on Fri Feb 9 19:25:07 2018 by

LaTeXML
proof that |g| divides exp⁡G

proof that

|

g

|

divides

exp

⁡

G

The following is a proof that, for every group

G

that has an exponent and for every

g

∈

G

,

|

g

|

divides

exp

⁡

G

.

Proof.

By the

division algorithm

, there exist

q

,

r

∈

ℤ

with

0

≤

r

<

|

g

|

such that

exp

⁡

G

=

q

⁢

|

g

|

+

r

. Since

e

G

=

g

exp

⁡

G

=

g

q

⁢

|

g

|

+

r

=

(

g

|

g

|

)

q

⁢

g

r

=

(

e

G

)

q

⁢

g

r

=

e

G

⁢

g

r

=

g

r

, by

definition

of the order of an element,

r

cannot be

positive

. Thus,

r

=

0

. It follows that

|

g

|

divides

exp

⁡

G

.
∎

Title

proof that

|

g

|

divides

exp

⁡

G

Canonical

name

ProofThatgDividesoperatornameexpG

Date of creation

2013-03-22 13:30:35

Last modified on

2013-03-22 13:30:35

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

8

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20D99

Generated on Fri Feb 9 19:25:10 2018 by

LaTeXML
proof that exp⁡G divides |G|

proof that

exp

⁡

G

divides

|

G

|

The following is a proof that

exp

⁡

G

divides

|

G

|

for every

finite group

G

.

Proof.

By the

division algorithm

, there exist

q

,

r

∈

ℤ

with

0

≤

r

<

exp

⁡

G

such that

|

G

|

=

q

⁢

(

exp

⁡

G

)

+

r

. Let

g

∈

G

. Then

e

G

=

g

|

G

|

=

g

q

⁢

(

exp

⁡

G

)

+

r

=

g

q

⁢

(

exp

⁡

G

)

⁢

g

r

=

(

g

exp

⁡

G

)

q

⁢

g

r

=

(

e

G

)

q

⁢

g

r

=

e

G

⁢

g

r

=

g

r

. Thus, for every

g

∈

G

,

g

r

=

e

G

. By the

definition

of

exponent

,

r

cannot be

positive

. Thus,

r

=

0

. It follows that

exp

⁡

G

divides

|

G

|

.
∎

Title

proof that

exp

⁡

G

divides

|

G

|

Canonical

name

ProofThatoperatornameexpGDividesG

Date of creation

2013-03-22 13:30:32

Last modified on

2013-03-22 13:30:32

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20D99

Generated on Fri Feb 9 19:25:13 2018 by

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subgroups with coprime orders

subgroups with coprime orders

If the orders of two

subgroups

of a group are coprime (

http://planetmath.org/Coprime

), the

identity element

is the only common

element of

the subgroups.

Proof.

Let

G

and

H

be such subgroups and

|

G

|

and

|

H

|

their orders.  Then the

intersection

G

∩

H

is a subgroup of both

G

and

H

.  By Lagrange’s theorem,

|

G

∩

H

|

divides both

|

G

|

and

|

H

|

and consequently it divides also

gcd

⁡

(

|

G

|

,

|

H

|

)

which is 1.  Therefore

|

G

∩

H

|

=

1

,  whence the intersection contains only the identity element.

Example.

All subgroups

{

(

1

)

,

(

12

)

}

,

{

(

1

)

,

(

13

)

}

,

{

(

1

)

,

(

23

)

}

of order 2 of the

symmetric group

𝔖

3

have only the identity element

(

1

)

common with the sole subgroup

{

(

1

)

,

(

123

)

,

(

132

)

}

of order 3.

Title

subgroups with coprime orders

Canonical name

SubgroupsWithCoprimeOrders

Date of creation

2013-03-22 18:55:58

Last modified on

2013-03-22 18:55:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 20D99

Related topic

Gcd

Related topic

CycleNotation

Generated on Fri Feb 9 19:25:15 2018 by

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Nielsen-Schreier theorem

Nielsen-Schreier theorem

Every

subgroup

(

http://planetmath.org/Subgroup

) of a

free group

is free.

This theorem is named after Jakob Nielsen, who in 1921 proved that every subgroup of a free group of finite rank (

http://planetmath.org/FreeGroup

) is free, and Otto Schreier, who proved the full result in 1927.

Title

Nielsen-Schreier theorem

Canonical name

NielsenSchreierTheorem

Date of creation

2013-03-22 12:54:07

Last modified on

2013-03-22 12:54:07

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

9

Author

yark (2760)

Entry type

Theorem

Classification

msc 20E05

Generated on Fri Feb 9 19:25:20 2018 by

LaTeXML
proof of Nielsen-Schreier theorem and Schreier index formula

proof of Nielsen-Schreier theorem and Schreier index formula

While there are purely algebraic proofs of the

Nielsen-Schreier theorem

, a much easier proof is available through geometric group theory.

Let

G

be a group which is free on a set

X

. Any group acts freely on its

Cayley graph

, and the Cayley graph of

G

is a

2

⁢

|

X

|

-

regular

tree, which we will call

𝒯

.

If

H

is any

subgroup

of

G

, then

H

also acts freely on

𝒯

by

restriction

. Since

groups that act freely on trees are free

,

H

is free.

Moreover, we can obtain the rank of

H

(the size of the set on which it is free). If

𝒢

is a

finite graph

, then

π

1

⁢

(

𝒢

)

is free of rank

-

χ

⁢

(

𝒢

)

-

1

, where

χ

⁢

(

𝒢

)

denotes the

Euler characteristic

of

𝒢

. Since

H

≅

π

1

⁢

(

H

\

𝒯

)

, the rank of

H

is

χ

⁢

(

H

\

𝒯

)

. If

H

is of finite index

n

in

G

, then

H

\

𝒯

is finite, and

χ

⁢

(

H

\

𝒯

)

=

n

⁢

χ

⁢

(

G

\

𝒯

)

. Of course

-

χ

⁢

(

G

\

𝒯

)

+

1

is the rank of

G

. Substituting, we obtain the

Schreier index formula

:

rank

⁢

(

H

)

=

n

⁢

(

rank

⁢

(

G

)

-

1

)

+

1

.

Title

proof of Nielsen-Schreier theorem and Schreier index formula

Canonical name

ProofOfNielsenSchreierTheoremAndSchreierIndexFormula

Date of creation

2013-03-22 13:56:02

Last modified on

2013-03-22 13:56:02

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Proof

Classification

msc 20E05

Classification

msc 20F65

Related topic

ScheierIndexFormula

Generated on Fri Feb 9 19:25:22 2018 by

LaTeXML
Schreier index formula

Schreier index formula

Let

F

be a

free group

of finite rank, and let

H

be a

subgroup

(

http://planetmath.org/Subgroup

) of finite index in

F

.
By the

Nielsen-Schreier theorem

,

H

is free.
The

Schreier index formula

states that

rank

(

H

)

=

|

F

:

H

|

⋅

(

rank

(

F

)

-

1

)

+

1

.

This implies more generally that if

G

is a group generated by

m

elements, then any subgroup of index

n

in

G

can be generated by at most

n

⁢

m

-

n

+

1

elements.

Title

Schreier index formula

Canonical name

SchreierIndexFormula

Date of creation

2013-03-22 13:56:18

Last modified on

2013-03-22 13:56:18

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

14

Author

yark (2760)

Entry type

Theorem

Classification

msc 20E05

Related topic

ProofOfNielsenSchreierTheoremAndSchreierIndexFormula

Generated on Fri Feb 9 19:25:28 2018 by

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free product

free product

Definition

Let

G

be a group, and let

(

A

i

)

i

∈

I

be
a family of

subgroups

(

http://planetmath.org/Subgroup

) of

G

.
Then

G

is said to be a

free product

of the subgroups

A

i

if given any group

H

and
a

homomorphism

(

http://planetmath.org/GroupHomomorphism

)

f

i

:

A

i

→

H

for each

i

∈

I

,
there is a unique homomorphism

f

:

G

→

H

such that

f

|

A

i

=

f

i

for all

i

∈

I

.
The subgroups

A

i

are then called the

free factors

of

G

.

If

G

is the free product of

(

A

i

)

i

∈

I

,
and

(

K

i

)

i

∈

I

is a family of groups such that

K

i

≅

A

i

for each

i

∈

I

,
then we may also say that

G

is the free product of

(

K

i

)

i

∈

I

.
With this definition, every family of groups has a free product,
and the free product is unique up to

isomorphism

.

The free product is the

coproduct

in the category of groups.

Construction

Free groups

are simply the free products of

infinite cyclic groups

,
and it is possible to generalize the construction given in the free group
article to the case of arbitrary free products.
But we will instead

construct

the free product as a

quotient

(

http://planetmath.org/QuotientGroup

) of a free group.

Let

(

K

i

)

i

∈

I

be a family of groups.
For each

i

∈

I

,
let

X

i

be a set and

γ

i

:

X

i

→

K

i

a

function

such that

γ

i

⁢

(

X

i

)

generates

K

i

.
The

X

i

should be chosen to be

pairwise disjoint

;
for example, we could take

X

i

=

K

i

×

{

i

}

,
and let

γ

i

be the obvious

bijection

.
Let

F

be a free group freely generated by

⋃

i

∈

I

X

i

.
For each

i

∈

I

,
the subgroup

⟨

X

i

⟩

of

F

is freely generated by

X

i

,
so there is a homomorphism

ϕ

i

:

⟨

X

i

⟩

→

K

i

extending

γ

i

.
Let

N

be the

normal closure

of

⋃

i

∈

I

ker

⁡

ϕ

i

in

F

.

Then it can be shown that

F

/

N

is
the free product of the family of subgroups

(

⟨

X

i

⟩

⁢

N

/

N

)

i

∈

I

,
and

K

i

≅

⟨

X

i

⟩

⁢

N

/

N

for each

i

∈

I

.

Title

free product

Canonical name

FreeProduct

Date of creation

2013-03-22 14:53:34

Last modified on

2013-03-22 14:53:34

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

12

Author

yark (2760)

Entry type

Definition

Classification

msc 20E06

Related topic

FreeProductWithAmalgamatedSubgroup

Related topic

FreeGroup

Defines

free factor

Generated on Fri Feb 9 19:25:30 2018 by

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