proof of fourth isomorphism theorem

proof of fourth isomorphism theorem

First we must prove that the map defined by

A

↦

A

/

N

is a

bijection

. Let

θ

denote this map, so that

θ

⁢

(

A

)

=

A

/

N

. Suppose

A

/

N

=

B

/

N

, then for any

a

∈

A

we have

a

⁢

N

=

b

⁢

N

for some

b

∈

B

, and so

b

-

1

⁢

a

∈

N

⊆

B

. Hence

A

⊆

B

, and similarly

B

⊆

A

, so

A

=

B

and

θ

is

injective

.
Now suppose

S

is a

subgroup

of

G

/

N

and

ϕ

:

G

→

G

/

N

by

ϕ

⁢

(

g

)

=

g

⁢

N

. Then

ϕ

-

1

⁢

(

S

)

=

{

s

∈

G

:

s

⁢

N

∈

S

}

is a subgroup of

G

containing

N

and

θ

⁢

(

ϕ

-

1

⁢

(

S

)

)

=

{

s

⁢

N

:

s

⁢

N

∈

S

}

=

S

, proving that

θ

is

bijective

.

Now we move to the given properties:

1.

A

≤

B

iff

A

/

N

≤

B

/

N

If

A

≤

B

then trivially

A

/

N

≤

B

/

N

, and the

converse

follows from the fact that

θ

is bijective.

2.

A

≤

B

implies

|

B

:

A

|

=

|

B

/

N

:

A

/

N

|

Let

ψ

map the cosets in

B

/

A

to the cosets in

(

B

/

N

)

/

(

A

/

N

)

by

mapping

the coset

b

⁢

A

b

∈

B

to the coset

(

b

⁢

N

)

⁢

(

A

/

N

)

. Then

ψ

is well defined and injective because:

b

1

⁢

A

=

b

2

⁢

A

⇔

b

1

-

1

⁢

b

2

∈

A

⇔

(

b

1

⁢

N

)

-

1

⁢

(

b

2

⁢

N

)

=

b

1

-

1

⁢

b

2

⁢

N

∈

A

/

N

⇔

(

b

1

⁢

N

)

⁢

(

A

/

N

)

=

(

b

2

⁢

N

)

⁢

(

A

/

N

)

.

Finally,

ψ

is

surjective

since

b

ranges over all of

B

in

(

b

⁢

N

)

⁢

(

A

/

N

)

.

3.

⟨

A

,

B

⟩

/

N

=

⟨

A

/

N

,

B

/

N

⟩

To show

⟨

A

,

B

⟩

/

N

⊆

⟨

A

/

N

,

B

/

N

⟩

we need only show that if

x

∈

A

or

x

∈

B

then

x

⁢

N

∈

⟨

A

/

N

,

B

/

N

⟩

. The other cases are dealt with using the fact that

(

x

⁢

y

)

⁢

N

=

(

x

⁢

N

)

⁢

(

y

⁢

N

)

. So suppose

x

∈

A

then clearly

x

⁢

N

∈

⟨

A

/

N

,

B

/

N

⟩

because

x

⁢

N

∈

A

/

N

. Similarly for

x

∈

B

.
Similarly, to show

⟨

A

/

N

,

B

/

N

⟩

⊆

⟨

A

,

B

⟩

/

N

we need only show that if

x

⁢

N

∈

A

/

N

or

x

⁢

N

∈

B

/

N

then

x

∈

⟨

A

,

B

⟩

. So suppose

x

⁢

N

∈

A

/

N

, then

x

⁢

N

=

a

⁢

N

for some

a

∈

A

, giving

a

-

1

⁢

x

∈

N

⊆

A

and so

x

∈

A

⊆

⟨

A

,

B

⟩

. Similarly for

x

⁢

N

∈

B

/

N

.

4.

(

A

∩

B

)

/

N

=

(

A

/

N

)

∩

(

B

/

N

)

Suppose

x

⁢

N

∈

(

A

∩

B

)

/

N

, then

x

⁢

N

=

y

⁢

N

for some

y

∈

(

A

∩

B

)

and since

N

⊆

(

A

∩

B

)

,

x

∈

(

A

∩

B

)

. Therefore

x

∈

A

and

x

∈

B

, and so

x

⁢

N

∈

(

A

/

N

)

∩

(

B

/

N

)

meaning

(

A

∩

B

)

/

N

⊆

(

A

/

N

)

∩

(

B

/

N

)

.
Now suppose

x

⁢

N

∈

(

A

/

N

)

∩

(

B

/

N

)

. Then

x

⁢

N

=

a

⁢

N

for some

a

∈

A

, giving

a

-

1

⁢

x

∈

N

⊆

A

and so

x

∈

A

. Similarly

x

∈

B

, therefore

x

⁢

N

∈

(

A

∩

B

)

/

N

and

(

A

/

N

)

∩

(

B

/

N

)

⊆

(

A

∩

B

)

/

N

.

5.

A

⊴

G

iff

(

A

/

N

)

⊴

(

G

/

N

)

Suppose

A

⊴

G

. Then for any

g

∈

G

we have

(

g

⁢

N

)

⁢

(

A

/

N

)

⁢

(

g

⁢

N

)

-

1

=

(

g

⁢

A

⁢

g

-

1

)

/

N

=

A

/

N

and so

(

A

/

N

)

⊴

(

G

/

N

)

.

Conversely suppose

(

A

/

N

)

⊴

(

G

/

N

)

. Consider

σ

:

g

↦

(

g

⁢

N

)

⁢

(

A

/

N

)

, the

composition

of the map from

G

onto

G

/

N

and the map from

G

/

N

onto

(

G

/

N

)

/

(

A

/

N

)

.

g

∈

ker

⁡

π

iff

(

g

⁢

N

)

⁢

(

A

/

N

)

=

(

A

/

N

)

which occurs iff

g

⁢

N

∈

A

/

N

therefore

g

⁢

N

=

a

⁢

N

for some

a

∈

A

. However

N

is

contained

in

A

, so this statement is equivalnet to saying

g

∈

A

. So

A

is the kernel of a

homomorphism

, hence is a

normal subgroup

of

G

.

Title

proof of fourth isomorphism theorem

Canonical name

ProofOfFourthIsomorphismTheorem

Date of creation

2013-03-22 14:17:38

Last modified on

2013-03-22 14:17:38

Owner

aoh45 (5079)

Last modified by

aoh45 (5079)

Numerical id

9

Author

aoh45 (5079)

Entry type

Proof

Classification

msc 20A05

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