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Alternating Group A3 is Simple

Normal Subgroups

And...

Trival Group (T)

Ts for every s in S3

Trivial subgroup (T):

1) A subgroup of S3

2) - sT = Ts for every s in S3

OR

- sTs^-1 = t for every s in S3

Alternating group A3 (A):

1) A subgroup of S3

2) - sA = As for every s in S3

OR

- sAs^-1 = A for every s in S3

A3 must not have any other Normal Subgroups

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

(1)

(1) (1)

(12) (12)

(13) (13)

(23) (23)

(123) (123)

(132) (132)

Trivial Group (T)

1) A group

2) sT=Ts for every s in S3

And...

If A3 is Simple...

A3 must be a Normal Subgroup

Trivial Group (T)

sT for every s in S3

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

Cayley Table of S3

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

Then...

A3 is a nontrivial group whose only normal subgroups are the trivial group and the group itself

That means...

The Trivial Group {(1)} must be a Normal Subgroup

Trivial Group (T)

(1)

(1) (1)

1) Closed

2) Associative

3) Identity

4) Inverses

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

What does A3 look like?

A3 = {(1) (123) (132)}

Alternating Group in A3 (A)

***We must show that sAs^-1 = A for every s in S3**

If the permutation always ends up being even, then it must be in the set of all even permutations which is A (A3).

A3 = {(1) (123) (132)}

Possible Subgroup:

{(1), (132)}

Not Closed -> Not a Group

Cannot be a Normal Subgroup

Alternating Group in A3 (A)

Alternating group A3 (A)

(1) (132)

(1) (1) (132)

(132) (132) (123)

Alternating Group in A3 (A)

Possible Subgroups of the Alternating group A3 (A)

**We must show that sAs^-1 = A for every s in S3**

Thus, A will still always be even.

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

Possible Subgroup:

{(1), (123)}

We have shown that sAs^-1 = A for every s in S3

**Normal Subgroup**

Not Closed -> Not a Group

Cannot be a Normal Subgroup

(1) (123)

(1) (1) (132)

(123) (132) (132)

1) A group

2) sAs^-1 = A for every s in S3

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

1) {(1), (123)}

2) {(1), (132)}

Alternating Group A3 (A)

(1) (123) (132)

(1) (1) (123) (132)

(123) (123) (132) (1)

(132) (132) (1) (123)

1) Closed

2) Associative

3) Identity

4) Inverses

Alternating Group in A3 (A)

**We must show that sAs^-1 = A for every s in S3**

Well... if s or s^-1 are ever operated on A they must both be odd or both be even...

(1) (12) (13) (23) (123) (132)

(1) (1) (12) (13) (23) (123) (132)

(12) (12) (1) (132) (123) (23) (13)

(13) (13) (123) (1) (132) (12) (23)

(23) (23) (132) (123) (1) (13) (12)

(123) (123) (13) (23) (12) (132) (1)

(132) (132) (23) (12) (13) (1) (123)

Alternating Group A3 (A)

Definition of Alternating Group A3:

The set of even permutations in S3.

That is,

(123) = (13)(12) -> 2 transpositions

(132) = (12)(13) -> 2 transpositions

(1) = (1) -> (can be constructed

with any combination

of an even number

of transpositions)

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