Discussion
Let \(S_{3}\) be the symmetric group, \(I_{n}(S_{3})\) be the group of inner automorphisms of \(S_{3}\) and \(Z(S_{3})\) be the center of \(S_{3}\). Then which is FALSE ?
Answer: C
The false statement is \(I_{n}(S_{3}) = \{e\}\).
The center of \(S_{3}\) is trivial: no non-identity permutation commutes with all others (for example (1 2) does not commute with (1 3)). So \(Z(S_{3}) = \{e\}\), the first option, is true.
The map \(g \mapsto\) (conjugation by \(g\)) is a homomorphism from \(S_{3}\) onto the inner automorphism group, with kernel the center. So \(I_{n}(S_{3}) \cong S_{3}/Z(S_{3})\), the last option, and since the center is trivial this is a copy of \(S_{3}\) itself, which the paper writes as \(I_{n}(S_{3}) = S_{3}\).
Consequently the inner automorphism group has 6 elements, not 1. It would be trivial only for an abelian group, where every conjugation is the identity map, and \(S_{3}\) is not abelian.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development