Find normal subgroups of S3
The correct answer and explanation is:
Correct Answer: The normal subgroups of S3S_3 are:
- {e}\{ e \} (the trivial subgroup)
- S3S_3 (the whole group)
- A3={e,(123),(132)}A_3 = \{ e, (123), (132) \}
Explanation:
The symmetric group S3S_3 is the group of all permutations of three elements. It has six elements: S3={e,(12),(13),(23),(123),(132)}S_3 = \{ e, (12), (13), (23), (123), (132) \}
A normal subgroup NN of a group GG satisfies gNg−1=NgNg^{-1} = N for all g∈Gg \in G. This means the subgroup is invariant under conjugation by any element of the group.
First, the trivial subgroup {e}\{ e \} and the whole group S3S_3 are always normal in any group.
To find other normal subgroups, examine the structure of S3S_3. It has the following subgroups:
- {e}\{ e \}
- {e,(12)}\{ e, (12) \}
- {e,(13)}\{ e, (13) \}
- {e,(23)}\{ e, (23) \}
- A3={e,(123),(132)}A_3 = \{ e, (123), (132) \}
- S3S_3
Check which of these are normal. The three subgroups of order 2 each contain one transposition. These are not normal. For example, conjugating (12)(12) by (13)(13) gives: (13)(12)(13)−1=(23)(13)(12)(13)^{-1} = (23)
This shows that {e,(12)}\{ e, (12) \} is not closed under conjugation, so it is not normal.
Now consider A3={e,(123),(132)}A_3 = \{ e, (123), (132) \}. It is of order 3. Since 3 divides 6, and the index of this subgroup is 2, which is the smallest nontrivial index, this subgroup is always normal. Any subgroup of index 2 in a group is normal because conjugation preserves the coset structure.
Therefore, the only normal subgroups of S3S_3 are:
- {e}\{ e \}
- A3A_3
- S3S_3
These subgroups are closed under conjugation and satisfy the definition of normality.