Permutations of a single number in the symmetric group

Click For Summary

Homework Help Overview

The discussion revolves around the symmetric group S_5 and the permutations of specific subsets, particularly focusing on the elements {2,5} and {3}. Participants explore the nature of identity permutations and the representation of cycles in permutations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the distinction between the identity permutation and the representation of 1-cycles. There is also a discussion about the number of permutations for different sets of elements.

Discussion Status

The conversation is ongoing, with participants clarifying concepts related to permutations and exploring the implications of representing identity permutations. Some guidance on the number of permutations based on the number of elements has been provided.

Contextual Notes

Participants are considering the conventions of writing permutations, particularly regarding 1-cycles and identity elements, as well as the factorial representation of permutations based on the number of elements.

Ted123
Messages
428
Reaction score
0
Say we have the symmetric group S_5.

The permutations of \{2,5\} are the identity e and the transposition (25).

But what are all the permutations of \{3\}? Is it e and the 1-cycle (3)?
 
Physics news on Phys.org
Ted123 said:
Say we have the symmetric group S_5.

The permutations of \{2,5\} are the identity e and the transposition (25).

But what are all the permutations of \{3\}? Is it e and the 1-cycle (3)?

What do you think would be the difference between e and (3)? Aren't they both the identity permutation?
 
Dick said:
What do you think would be the difference between e and (3)? Aren't they both the identity permutation?

Oh yeah of course. You don't have to write 1-cycles in a permutation so e=(1)(2)(3)(4)(5)=(1)=(2)=(3)=(4)=(5)=(1)(2)=(1)(3) etc.
 
Ted123 said:
Oh yeah of course. You don't have to write 1-cycles in a permutation so e=(1)(2)(3)(4)(5)=(1)=(2)=(3)=(4)=(5)=(1)(2)=(1)(3) etc.

Right. The number of permutations on n elements is n!. So for 2 elements you have two permutations, for 1 element you have one.
 

Similar threads

Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
1K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
6K