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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)
1) A group
2) sT=Ts for every s in S3
A3 must be a Normal Subgroup
sT for every s in S3
(1) (12) (13) (23) (123) (132)
(1) (1) (12) (13) (23) (123) (132)
(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
The Trivial Group {(1)} must be a Normal Subgroup
(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)
A3 = {(1) (123) (132)}
***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)}
Not Closed -> Not a Group
Cannot be a Normal Subgroup
(1) (132)
(1) (1) (132)
(132) (132) (123)
**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)
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)}
(1) (123) (132)
(1) (1) (123) (132)
(123) (123) (132) (1)
(132) (132) (1) (123)
1) Closed
2) Associative
3) Identity
4) Inverses
**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)
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)