Proving ##\frac{n!}{r(n-r)!}## Distinct r-Cycles in $S_n$

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SUMMARY

In the symmetric group \( S_n \), there are exactly \( \frac{n!}{r(n-r)!} \) distinct r-cycles. This formula arises from the need to account for the cyclic nature of r-cycles, where two cycles are considered identical if they are cyclic permutations of each other. By fixing one object at the start of the cycle, the remaining \( r-1 \) objects can be permuted freely, leading to the distinct arrangements. The initial confusion stemmed from the incorrect application of the binomial coefficient \( \frac{n!}{r!(n-r)!} \) without considering the cyclic permutations.

PREREQUISITES
  • Understanding of symmetric groups, specifically \( S_n \)
  • Familiarity with permutations and combinations
  • Knowledge of cycle notation in group theory
  • Basic grasp of factorial notation and its applications
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  • Learn about the concept of cycle notation in group theory
  • Explore combinatorial proofs involving permutations and combinations
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Mathematics students, particularly those studying abstract algebra, combinatorics, or group theory, will benefit from this discussion.

Lee33
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Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}## but I don't know how they got ##\frac{n!}{r(n-r)!}##?
 
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Lee33 said:

Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}##

Two r-cycles are the same if their cycle notations are cyclic permutations of each other. Having chosen our objects, we can avoid such multiple counting by fixing one object to appear the beginning of all the r-cycles. Each permutation of the remaining [itex]r-1[/itex] objects will then generate a distinct r-cycle.
 

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