aarjaneiro/planetmath huggingface dataset:
presentation of a group

presentation of a group

A

presentation

of a group

G

is a description of

G

in terms of
generators and relations (sometimes also known as relators).
We say that the group is finitely
presented, if it can be described in terms of a finite number of

generators

and a finite number of

defining relations

. A

collection

of
group

elements

g

i

∈

G

,

i

∈

I

is said to generate

G

if every
element of

G

can be specified as a

product

of the

g

i

, and of their

inverses

. A

relation

is a word over the

alphabet

consisting of the
generators

g

i

and their inverses, with the

property

that it
multiplies out to the

identity

in

G

. A set of relations

r

j

,

j

∈

J

is said to be defining, if all relations in

G

can be given
as a product of the

r

j

, their inverses, and the

G

-

conjugates

of
these.

The standard notation for the presentation of a group is

G

=

⟨

g

i

∣

r

j

⟩

,

meaning that

G

is generated by generators

g

i

, subject to
relations

r

j

. Equivalently, one has a

short exact sequence

of
groups

1

→

N

→

F

⁢

[

I

]

→

G

→

1

,

where

F

⁢

[

I

]

denotes the

free group

generated by the

g

i

, and where

N

is the

smallest normal subgroup
containing

all the

r

j

. By the

Nielsen-Schreier Theorem

, the kernel

N

is itself a free group, and hence we assume without loss of generality
that there are no relations among the relations.

Example.

The

symmetric group

on

n

elements

1

,

…

,

n

admits the following finite presentation (Note: this presentation is
not

canonical

. Other presentations are known.) As generators take

g

i

=

(

i

,

i

+

1

)

,

i

=

1

,

…

,

n

-

1

,

the

transpositions

of

adjacent

elements. As defining relations take

(

g

i

⁢

g

j

)

n

i

,

j

=

id

,

i

,

j

=

1

,

…

⁢

n

,

where

n

i

,

i

=

1

n

i

,

i

+

1

=

3

n

i

,

j

=

2

,

|

j

-

i

|

>

1

.

This means that a finite symmetric group is a

Coxeter group

.

Title

presentation of a group

Canonical name

PresentationOfAGroup

Date of creation

2013-03-22 12:23:23

Last modified on

2013-03-22 12:23:23

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

20

Author

rmilson (146)

Entry type

Definition

Classification

msc 20A05

Classification

msc 20F05

Synonym

presentation

Synonym

finite presentation

Synonym

finitely presented

Related topic

GeneratingSetOfAGroup

Related topic

CayleyGraph

Defines

generator

Defines

relation

Defines

generators and relations

Defines

relator

Generated on Fri Feb 9 19:18:29 2018 by

LaTeXML
proof of first isomorphism theorem

proof of first isomorphism theorem

The proof consist of several parts which we will give for
completeness. Let

K

denote

ker

⁡

f

. The following calculation
validates that for every

g

∈

G

and

k

∈

K

:

f

⁢

(

g

⁢

k

⁢

g

-

1

)

=

f

⁢

(

g

)

⁢

f

⁢

(

k

)

⁢

f

⁢

(

g

)

-

1

(

f

is an homomorphism)

=

f

⁢

(

g

)

⁢

1

H

⁢

f

⁢

(

g

)

-

1

(definition of

K

)

=

1

H

Hence,

g

⁢

k

⁢

g

-

1

is in

K

. Therefore,

K

is a

normal subgroup

of

G

and

G

/

K

is

well-defined

.

To prove the

theorem

we will define a map from

G

/

K

to the image
of

f

and show that it is a

function

, a

homomorphism

and finally
an

isomorphism

.

Let

θ

:

G

/

K

→

Im

⁡

f

be a map that sends the coset

g

⁢

K

to

f

⁢

(

g

)

.

Since

θ

is defined on representatives we need to show that
it is well defined. So, let

g

1

and

g

2

be two elements of

G

that belong to the same coset (i.e.

g

1

⁢

K

=

g

2

⁢

K

). Then,

g

1

-

1

⁢

g

2

is an element of

K

and therefore

f

⁢

(

g

1

-

1

⁢

g

2

)

=

1

(because

K

is the kernel of

G

). Now, the
rules of homomorphism show that

f

⁢

(

g

1

)

-

1

⁢

f

⁢

(

g

2

)

=

1

and that is

equivalent

to

f

⁢

(

g

1

)

=

f

⁢

(

g

2

)

which implies the equality

θ

⁢

(

g

1

⁢

K

)

=

θ

⁢

(

g

2

⁢

K

)

.

Next we verify that

θ

is a homomorphism. Take two cosets

g

1

⁢

K

and

g

2

⁢

K

, then:

θ

⁢

(

g

1

⁢

K

⋅

g

2

⁢

K

)

=

θ

⁢

(

g

1

⁢

g

2

⁢

K

)

(operation in

G

/

K

)

=

f

⁢

(

g

1

⁢

g

2

)

(definition of

θ

)

=

f

⁢

(

g

1

)

⁢

f

⁢

(

g

2

)

(

f

is an homomorphism)

=

θ

⁢

(

g

1

⁢

K

)

⁢

θ

⁢

(

g

2

⁢

K

)

(definition of

θ

)

Finally, we show that

θ

is an isomorphism (i.e. a

bijection

). The kernel of

θ

consists of all cosets

g

⁢

K

in

G

/

K

such that

f

⁢

(

g

)

=

1

but these are exactly the

elements

g

that belong to

K

so only the coset

K

is in the kernel of

θ

which implies that

θ

is an

injection

. Let

h

be an element of

Im

⁡

f

and

g

its pre-image. Then,

θ

⁢

(

g

⁢

K

)

equals

f

⁢

(

g

)

thus

θ

⁢

(

g

⁢

K

)

=

h

and therefore

θ

is

surjective

.

The theorem is proved. Some version of the theorem also states that
the following

diagram

is

commutative

:

\xymatrix

⁢

G

⁢

\ar

⁢

[

r

⁢

d

]

f

⁢

\ar

⁢

[

r

]

π

⁢

&

⁢

G

/

K

⁢

\ar

⁢

[

d

]

θ

⁢

&

⁢

H

were

π

is the

natural projection

that takes

g

∈

G

to

g

⁢

K

.
We will conclude by verifying this. Take

g

in

G

then,

θ

⁢

(

π

⁢

(

g

)

)

=

θ

⁢

(

g

⁢

K

)

=

f

⁢

(

g

)

as needed.

Title

proof of first isomorphism theorem

Canonical

name

ProofOfFirstIsomorphismTheorem

Date of creation

2013-03-22 12:39:19

Last modified on

2013-03-22 12:39:19

Owner

uriw (288)

Last modified by

uriw (288)

Numerical id

9

Author

uriw (288)

Entry type

Proof

Classification

msc 20A05

Generated on Fri Feb 9 19:18:32 2018 by

LaTeXML
proof of second isomorphism theorem for groups

proof of second isomorphism theorem for groups

First, we shall prove that

H

⁢

K

is a

subgroup

of

G

:
Since

e

∈

H

and

e

∈

K

, clearly

e

=

e

2

∈

H

⁢

K

.
Take

h

1

,

h

2

∈

H

,

k

1

,

k

2

∈

K

.
Clearly

h

1

⁢

k

1

,

h

2

⁢

k

2

∈

H

⁢

K

.
Further,

h

1

⁢

k

1

⁢

h

2

⁢

k

2

=

h

1

⁢

(

h

2

⁢

h

2

-

1

)

⁢

k

1

⁢

h

2

⁢

k

2

=

h

1

⁢

h

2

⁢

(

h

2

-

1

⁢

k

1

⁢

h

2

)

⁢

k

2

Since

K

is a

normal subgroup

of

G

and

h

2

∈

G

,
then

h

2

-

1

⁢

k

1

⁢

h

2

∈

K

.
Therefore

h

1

⁢

h

2

⁢

(

h

2

-

1

⁢

k

1

⁢

h

2

)

⁢

k

2

∈

H

⁢

K

,
so

H

⁢

K

is

closed under

multiplication

.

Also,

(

h

⁢

k

)

-

1

∈

H

⁢

K

for

h

∈

H

,

k

∈

K

, since

(

h

⁢

k

)

-

1

=

k

-

1

⁢

h

-

1

=

h

-

1

⁢

h

⁢

k

-

1

⁢

h

-

1

and

h

⁢

k

-

1

⁢

h

-

1

∈

K

since

K

is a normal subgroup of

G

.
So

H

⁢

K

is closed under

inverses

, and is thus a subgroup of

G

.

Since

H

⁢

K

is a subgroup of

G

,
the normality of

K

in

H

⁢

K

follows immediately from
the normality of

K

in

G

.

Clearly

H

∩

K

is a subgroup of

G

,
since it is the

intersection

of two subgroups of

G

.

Finally, define

ϕ

:

H

→

H

⁢

K

/

K

by

ϕ

⁢

(

h

)

=

h

⁢

K

.
We claim that

ϕ

is a

surjective

homomorphism

from

H

to

H

⁢

K

/

K

.
Let

h

0

⁢

k

0

⁢

K

be some element of

H

⁢

K

/

K

;
since

k

0

∈

K

, then

h

0

⁢

k

0

⁢

K

=

h

0

⁢

K

, and

ϕ

⁢

(

h

0

)

=

h

0

⁢

K

.
Now

ker

⁡

(

ϕ

)

=

{

h

∈

H

∣

ϕ

⁢

(

h

)

=

K

}

=

{

h

∈

H

∣

h

⁢

K

=

K

}

and if

h

⁢

K

=

K

, then we must have

h

∈

K

. So

ker

⁡

(

ϕ

)

=

{

h

∈

H

∣

h

∈

K

}

=

H

∩

K

Thus, since

ϕ

⁢

(

H

)

=

H

⁢

K

/

K

and

ker

⁡

ϕ

=

H

∩

K

,
by the

First Isomorphism Theorem

we see that

H

∩

K

is normal in

H

and that there is a canonical isomorphism between

H

/

(

H

∩

K

)

and

H

⁢

K

/

K

.

Title

proof of second isomorphism theorem for groups

Canonical

name

ProofOfSecondIsomorphismTheoremForGroups

Date of creation

2013-03-22 12:49:47

Last modified on

2013-03-22 12:49:47

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

17

Author

yark (2760)

Entry type

Proof

Classification

msc 20A05

Related topic

ProofOfSecondIsomorphismTheoremForRings

Generated on Fri Feb 9 19:18:40 2018 by

LaTeXML
proof of third isomorphism theorem

proof of third isomorphism theorem

We’ll give a proof of the

third isomorphism theorem

using the

Fundamental homomorphism theorem

.

Let

G

be a group, and let

K

⊆

H

be

normal subgroups

of

G

. Define

p

,

q

to be the

natural homomorphisms

from

G

to

G

/

H

,

G

/

K

respectively:

p

⁢

(

g

)

=

g

⁢

H

,

q

⁢

(

g

)

=

g

⁢

K

⁢

∀

g

∈

G

.

K

is a subset of

ker

⁡

(

p

)

, so there exists a unique

homomorphism

φ

:

G

/

K

→

G

/

H

so that

φ

∘

q

=

p

.

p

is

surjective

, so

φ

is surjective as well; hence

im

⁡

φ

=

G

/

H

. The kernel of

φ

is

ker

⁡

(

p

)

/

K

=

H

/

K

. So by the

first isomorphism theorem

we have

(

G

/

K

)

/

ker

⁡

(

φ

)

=

(

G

/

K

)

/

(

H

/

K

)

≈

im

⁡

φ

=

G

/

H

.

Title

proof of third isomorphism theorem

Canonical name

ProofOfThirdIsomorphismTheorem

Date of creation

2013-03-22 15:35:09

Last modified on

2013-03-22 15:35:09

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

5

Author

Thomas Heye (1234)

Entry type

Proof

Classification

msc 20A05

Generated on Fri Feb 9 19:18:42 2018 by

LaTeXML
proof that all cyclic groups are abelian

proof that all cyclic groups are abelian

The following is a

proof that all cyclic groups are abelian

.

Proof.

Let

G

be a

cyclic group

and

g

be a

generator

of

G

. Let

a

,

b

∈

G

. Then there exist

x

,

y

∈

ℤ

such that

a

=

g

x

and

b

=

g

y

. Since

a

⁢

b

=

g

x

⁢

g

y

=

g

x

+

y

=

g

y

+

x

=

g

y

⁢

g

x

=

b

⁢

a

, it follows that

G

is

abelian

.
∎

Title

proof that all cyclic groups are abelian

Canonical name

ProofThatAllCyclicGroupsAreAbelian

Date of creation

2013-03-22 13:30:44

Last modified on

2013-03-22 13:30:44

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20A05

Generated on Fri Feb 9 19:18:45 2018 by

LaTeXML
proof that all cyclic groups of the same order are isomorphic to each other

proof that all cyclic groups of the same order are isomorphic to each other

The following is a proof that all

cyclic groups

of the same order are

isomorphic

to each other.

Proof.

Let

G

be a cyclic group and

g

be a

generator

of

G

. Define

φ

:

ℤ

→

G

by

φ

⁢

(

c

)

=

g

c

. Since

φ

⁢

(

a

+

b

)

=

g

a

+

b

=

g

a

⁢

g

b

=

φ

⁢

(

a

)

⁢

φ

⁢

(

b

)

,

φ

is a

group homomorphism

. If

h

∈

G

, then there exists

x

∈

ℤ

such that

h

=

g

x

. Since

φ

⁢

(

x

)

=

g

x

=

h

,

φ

is

surjective

.

Note that

ker

⁡

φ

=

{

c

∈

ℤ

:

φ

⁢

(

c

)

=

e

G

}

=

{

c

∈

ℤ

:

g

c

=

e

G

}

.

If

G

is

infinite

, then

ker

⁡

φ

=

{

0

}

, and

φ

is

injective

. Hence,

φ

is a group isomorphism, and

G

≅

ℤ

.

If

G

is finite, then let

|

G

|

=

n

. Thus,

|

g

|

=

|

⟨

g

⟩

|

=

|

G

|

=

n

. If

g

c

=

e

G

, then

n

divides

c

. Therefore,

ker

⁡

φ

=

n

⁢

ℤ

. By the

first isomorphism theorem

,

G

≅

ℤ

/

n

⁢

ℤ

=

ℤ

n

.

Let

H

and

K

be cyclic groups of the same order. If

H

and

K

are infinite, then, by the above ,

H

≅

ℤ

and

K

≅

ℤ

. If

H

and

K

are finite of order

n

, then, by the above ,

H

≅

ℤ

n

and

K

≅

ℤ

n

. In any case, it follows that

H

≅

K

.
∎

Title

proof that all cyclic groups of the same order are isomorphic to each other

Canonical name

ProofThatAllCyclicGroupsOfTheSameOrderAreIsomorphicToEachOther

Date of creation

2013-03-22 13:30:41

Last modified on

2013-03-22 13:30:41

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

9

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20A05

Generated on Fri Feb 9 19:18:48 2018 by

LaTeXML
proof that all subgroups of a cyclic group are cyclic

proof that all subgroups of a cyclic group are cyclic

The following is a proof that all

subgroups

of a

cyclic group

are cyclic.

Proof.

Let

G

be a cyclic group and

H

≤

G

. If

G

is trivial, then

H

=

G

, and

H

is cyclic. If

H

is the trivial subgroup, then

H

=

{

e

G

}

=

⟨

e

G

⟩

, and

H

is cyclic. Thus, for the of the proof, it will be assumed that both

G

and

H

are nontrivial.

Let

g

be a

generator

of

G

. Let

n

be the smallest

positive integer

such that

g

n

∈

H

.

Claim:

H

=

⟨

g

n

⟩

Let

a

∈

⟨

g

n

⟩

. Then there exists

z

∈

ℤ

with

a

=

(

g

n

)

z

. Since

g

n

∈

H

, we have that

(

g

n

)

z

∈

H

. Thus,

a

∈

H

. Hence,

⟨

g

n

⟩

⊆

H

.

Let

h

∈

H

. Then

h

∈

G

. Let

x

∈

ℤ

with

h

=

g

x

. By the

division algorithm

, there exist

q

,

r

∈

ℤ

with

0

≤

r

<

n

such that

x

=

q

⁢

n

+

r

. Thus,

h

=

g

x

=

g

q

⁢

n

+

r

=

g

q

⁢

n

⁢

g

r

=

(

g

n

)

q

⁢

g

r

. Therefore,

g

r

=

h

⁢

(

g

n

)

-

q

. Recall that

h

,

g

n

∈

H

. Hence,

g

r

∈

H

. By choice of

n

,

r

cannot be

positive

. Thus,

r

=

0

. Therefore,

h

=

(

g

n

)

q

⁢

g

0

=

(

g

n

)

q

⁢

e

G

=

(

g

n

)

q

∈

⟨

g

n

⟩

. Hence,

H

⊆

⟨

g

n

⟩

.

This proves the claim. It follows that every subgroup of

G

is cyclic.
∎

Title

proof that all subgroups of a cyclic group are cyclic

Canonical

name

ProofThatAllSubgroupsOfACyclicGroupAreCyclic

Date of creation

2013-03-22 13:30:47

Last modified on

2013-03-22 13:30:47

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 20A05

Related topic

ProofThatEverySubringOfACyclicRingIsACyclicRing

Generated on Fri Feb 9 19:18:51 2018 by

LaTeXML
proof that group homomorphisms preserve identity

proof that group homomorphisms preserve identity

Theorem

.

A

group homomorphism

preserves

identity elements

.

Proof.

Let

ϕ

:

G

→

K

be a group homomorphism.
For clarity we use

∗

and

⋆

for the

group operations

of

G

and

K

, respectively. Also,
denote the

identities

by

1

G

and

1

H

respectively.

By the

definition

of identity,

1

G

∗

1

G

=

1

G

.

(1)

Applying the

homomorphism

ϕ

to (

1

) produces:

ϕ

⁢

(

1

G

)

⋆

ϕ

⁢

(

1

G

)

=

ϕ

⁢

(

1

G

∗

1

G

)

=

ϕ

⁢

(

1

G

)

.

(2)

Multiply both sides of (

2

) by the

inverse

of

ϕ

⁢

(

1

G

)

in

K

,
and use the

associativity

of

⋆

to produce:

ϕ

⁢

(

1

G

)

=

(

ϕ

⁢

(

1

G

)

)

-

1

⋆

ϕ

⁢

(

1

G

)

⋆

ϕ

⁢

(

1

G

)

=

(

ϕ

⁢

(

1

G

)

)

-

1

⋆

ϕ

⁢

(

1

G

)

=

1

K

.

(3)

∎

Title

proof that group homomorphisms preserve identity

Canonical

name

ProofThatGroupHomomorphismsPreserveIdentity

Date of creation

2013-11-16 4:44:43

Last modified on

2013-11-16 4:44:43

Owner

jacou (1000048)

Last modified by

(0)

Numerical id

9

Author

jacou (0)

Entry type

Proof

Classification

msc 20A05

Synonym

1234

Related topic

1234

Defines

1234

Generated on Fri Feb 9 19:18:53 2018 by

LaTeXML
Proof: The orbit of any element of a group is a subgroup

Proof: The orbit of any element of a group is a subgroup

Following is a proof that, if

G

is a group and

g

∈

G

, then

⟨

g

⟩

≤

G

. Here

⟨

g

⟩

is the orbit of

g

and is defined as

⟨

g

⟩

=

{

g

n

:

n

∈

ℤ

}

Since

g

∈

⟨

g

⟩

, then

⟨

g

⟩

is nonempty.

Let

a

,

b

∈

⟨

g

⟩

. Then there exist

x

,

y

∈

ℤ

such that

a

=

g

x

and

b

=

g

y

. Since

a

⁢

b

-

1

=

g

x

⁢

(

g

y

)

-

1

=

g

x

⁢

g

-

y

=

g

x

-

y

∈

⟨

g

⟩

, it follows that

⟨

g

⟩

≤

G

.

Title

Proof: The orbit of any element of a group is a

subgroup

Canonical name

ProofTheOrbitOfAnyElementOfAGroupIsASubgroup

Date of creation

2013-03-22 13:30:58

Last modified on

2013-03-22 13:30:58

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Proof

Classification

msc 20A05

Related topic

Group

Related topic

Subgroup

Related topic

ProofThatEveryGroupOfPrimeOrderIsCyclic

Related topic

ProofOfTheConverseOfLagrangesTheoremForCyclicGroups

Defines

orbit

Generated on Fri Feb 9 19:18:55 2018 by

LaTeXML
properties of conjugacy

properties of conjugacy

Let

S

be a nonempty subset of a group

G

.  When

g

is an element of

G

, a

conjugate

of

S

is the subset

g

S

g

-

1

=

{

g

s

g

-

1

⋮

s

∈

S

}

.

We denote here

g

⁢

S

⁢

g

-

1

:=

S

g

.

(1)

If

T

is another nonempty subset and

h

another element of

G

, then it’s easily verified the formulae

•

(

S

⁢

T

)

g

=

S

g

⁢

T

g

•

(

S

g

)

h

=

S

g

⁢

h

The conjugates

H

g

of a

subgroup

H

of

G

are subgroups of

G

, since any mapping

x

↦

g

⁢

x

⁢

g

-

1

is an

automorphism

(an

inner automorphism

) of

G

and the

homomorphic image of group

is always a group.

The notation (1) can be extended to

⟨

S

g

⋮

g

∈

G

⟩

:=

S

G

(2)

where the angle parentheses express a generated subgroup.

S

G

is the least

normal subgroup

of

G

containing the subset

S

, and it is called the

normal closure

of

S

.

http://en.wikipedia.org/wiki/Conjugacy

Wiki

Title

properties of conjugacy

Canonical name

PropertiesOfConjugacy

Date of creation

2013-03-22 18:56:35

Last modified on

2013-03-22 18:56:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Topic

Classification

msc 20A05

Related topic

NormalClosure2

Related topic

NonIsomorphicGroupsOfGivenOrder

Defines

normal closure

Generated on Fri Feb 9 19:18:58 2018 by

LaTeXML
regular group action

regular group action

Let

G

be a

group action

on a set

X

.
The action is called if for any pair

α

,

β

∈

X

there
exists exactly one

g

∈

G

such that

g

⋅

α

=

β

. (For a
right
group action it is defined correspondingly.)

A key example of a

regular

action is the

regular representation

of a group, with action given by group

multiplication

.

Title

regular group action

Canonical

name

RegularGroupAction

Date of creation

2013-03-22 13:21:35

Last modified on

2013-03-22 13:21:35

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 20A05

Related topic

GroupAction

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

LaTeXML
solvable group

solvable group

A group

G

is

solvable

if it has a

subnormal series

G

=

G

0

⊃

G

1

⊃

⋯

⊃

G

n

=

{

1

}

where all the

quotient groups

G

i

/

G

i

+

1

are

abelian

.

Title

solvable group

Canonical name

SolvableGroup

Date of creation

2013-03-22 12:08:53

Last modified on

2013-03-22 12:08:53

Owner

djao (24)

Last modified by

djao (24)

Numerical id

11

Author

djao (24)

Entry type

Definition

Classification

msc 20A05

Synonym

solvable

Synonym

soluble group

Related topic

DerivedSubgroup

Related topic

CompositionSeries2

Related topic

GaloisCriterionForSolvabilityOfAPolynomialByRadicals

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

LaTeXML
subgroups of finite cyclic group

subgroups of finite cyclic group

Let

n

be the order of a finite

cyclic group

G

.  For every positive divisor (

http://planetmath.org/Divisibility

)

m

of

n

, there exists one and only one

subgroup

of order

m

of

G

.  The group

G

has no other subgroups.

Proof.

If

g

is a

generator

of

G

and

n

=

m

⁢

k

,  then

g

k

generates the subgroup

⟨

g

k

⟩

, the order of which is equal to the order of

g

k

, i.e. equal to

m

.  Any subgroup

H

of

G

is cyclic (see

http://planetmath.org/node/4097

this entry).  If

|

H

|

=

m

,  then

H

must have a generator of order

m

; thus apparently

H

=

⟨

g

±

k

⟩

=

⟨

g

k

⟩

.

Title

subgroups of finite cyclic group

Canonical name

SubgroupsOfFiniteCyclicGroup

Date of creation

2013-03-22 18:57:13

Last modified on

2013-03-22 18:57:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 20A05

Generated on Fri Feb 9 19:19:12 2018 by

LaTeXML
the derived subgroup is normal

the derived subgroup is normal

We are going to prove:

”The

derived subgroup

(or

commutator subgroup

)

[

G

,

G

]

is normal in

G

”

Proof:

We have to show that for each

x

∈

[

G

,

G

]

,

g

⁢

x

⁢

g

-

1

it is also in

[

G

,

G

]

.

Since

[

G

,

G

]

is the

subgroup

generated by the all commutators in

G

, then for each

x

∈

[

G

,

G

]

we have

x

=

c

1

⁢

c

2

⁢

⋯

⁢

c

m

–a word of commutators– so

c

i

=

[

a

i

,

b

i

]

for all

i

.

Now taking any element of

g

∈

G

we can see that

g

⁢

[

a

i

,

b

i

]

⁢

g

-

1

=

g

⁢

a

i

⁢

b

i

⁢

a

i

-

1

⁢

b

i

-

1

⁢

g

-

1

=

g

⁢

a

i

⁢

g

-

1

⁢

g

⁢

b

i

⁢

g

-

1

⁢

g

⁢

a

i

-

1

⁢

g

-

1

⁢

g

⁢

b

i

-

1

⁢

g

-

1

=

(

g

⁢

a

i

⁢

g

-

1

)

⁢

(

g

⁢

b

i

⁢

g

-

1

)

⁢

(

g

⁢

a

i

⁢

g

-

1

)

-

1

⁢

(

g

⁢

b

i

⁢

g

-

1

)

-

1

=

[

g

⁢

a

i

⁢

g

-

1

,

g

⁢

b

i

⁢

g

-

1

]

,

that is

g

⁢

[

a

i

,

b

i

]

⁢

g

-

1

=

[

g

⁢

a

i

⁢

g

-

1

,

g

⁢

b

i

⁢

g

-

1

]

so a

conjugation

of a commutator is another commutator, then
for the conjugation

g

⁢

x

⁢

g

-

1

=

g

⁢

c

1

⁢

c

2

⁢

⋯

⁢

c

m

⁢

g

-

1

=

g

⁢

c

1

⁢

g

-

1

⁢

g

⁢

c

2

⁢

g

-

1

⁢

g

⁢

⋯

⁢

g

-

1

⁢

g

⁢

c

m

⁢

g

-

1

=

(

g

⁢

c

1

⁢

g

-

1

)

⁢

(

g

⁢

c

2

⁢

g

-

1

)

⁢

⋯

⁢

(

g

⁢

c

m

⁢

g

-

1

)

is another word of commutators, hence

g

⁢

x

⁢

g

-

1

is in

[

G

,

G

]

which in turn implies that

[

G

,

G

]

is normal in

G

, QED.

Title

the

derived subgroup is normal

Canonical name

TheDerivedSubgroupIsNormal

Date of creation

2013-03-22 16:04:39

Last modified on

2013-03-22 16:04:39

Owner

juanman (12619)

Last modified by

juanman (12619)

Numerical id

8

Author

juanman (12619)

Entry type

Proof

Classification

msc 20A05

Classification

msc 20E15

Classification

msc 20F14

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

LaTeXML
The History of Having Settled To Accomplish Studies

The History of Having Settled To Accomplish Studies

Theorem

A

group homomorphism

preserves inverses elements. That is, for groups

(

G

,

∗

)

and

(

K

,

⋆

)

, and a

homomorphism

ϕ

:

G

→

K

,

ϕ

⁢

(

x

-

1

)

=

ϕ

⁢

(

x

)

-

1

.

{proof}

Fix an

x

∈

G

. Observe that

ϕ

⁢

(

x

∗

x

-

1

)

=

ϕ

⁢

(

1

G

)

=

ϕ

⁢

(

x

)

⋆

ϕ

⁢

(

x

-

1

)

(1)

Recall that, for any group homomorphism

ϕ

:

G

→

K

,

ϕ

⁢

(

1

G

)

=

1

K

(2)

In other , homomorphisms preserve

identity

.

1

1

A proof for that statement is attached to the .

It follows from (

1

) and (

2

) that

ϕ

⁢

(

x

)

⋆

ϕ

⁢

(

x

-

1

)

=

1

K

(3)

Because the

inverse

of any group is unique, the only value of

ϕ

⁢

(

x

-

1

)

whose

product

with

ϕ

⁢

(

x

)

is

1

K

is, of course,

ϕ

⁢

(

x

)

-

1

. Therefore, all group homomorphisms preserve the inverse.

Title

The History of Having Settled To Accomplish Studies

Canonical name

TheHistoryOfHavingSettledToAccomplishStudies

Date of creation

2013-11-27 10:59:44

Last modified on

2013-11-27 10:59:44

Owner

jacou (1000048)

Last modified by

(0)

Numerical id

14

Author

jacou (0)

Entry type

Proof

Classification

msc 20A05

Generated on Fri Feb 9 19:19:18 2018 by

LaTeXML