Permutations of a single number in the symmetric group

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SUMMARY

The discussion focuses on the permutations within the symmetric group S_5, specifically examining the permutations of the set {2,5} and the single element {3}. It is established that the permutations of {2,5} include the identity permutation e and the transposition (25). For the single element {3}, the only permutations are the identity e and the 1-cycle (3), which is equivalent to e in notation but not in function. The total number of permutations for n elements is confirmed to be n!, illustrating that for 2 elements there are 2 permutations and for 1 element there is 1 permutation.

PREREQUISITES
  • Understanding of symmetric groups, specifically S_n
  • Familiarity with permutation notation and concepts
  • Knowledge of identity permutations and cycles
  • Basic combinatorial principles, particularly factorials
NEXT STEPS
  • Study the properties of symmetric groups, focusing on S_n
  • Learn about cycle notation in permutations
  • Explore the concept of transpositions and their role in permutation groups
  • Investigate combinatorial proofs involving factorials and permutations
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone interested in group theory and combinatorial mathematics will benefit from this discussion.

Ted123
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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)?
 
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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.
 

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