Number of Elements of Order 5 in S7 Permutation Group

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Discussion Overview

The discussion revolves around determining the number of elements of order 5 in the permutation group S7. Participants explore the implications of cycle decomposition, particularly focusing on the least common multiple (l.c.m) of cycle lengths and the arrangement of elements within those cycles.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that to achieve an l.c.m of 5, two elements must be fixed, leading to a calculation of 7C2 * 4! = 504, but later questions the validity of this result.
  • Another participant asserts that since 5 is prime, a cycle of length 5 must be present, and a 2-cycle could be considered, although this would result in an l.c.m of 10, which is not acceptable for this case.
  • It is noted that the only permutations of order 5 in S7 are 5-cycles.
  • A participant points out that the initial calculation incorrectly counts distinct arrangements of the same cycle as different elements.
  • There is a suggestion to divide by 4 to account for equivalent cycle arrangements.
  • Another participant provides a combinatorial approach, stating that there are {7 choose 5} ways to select 5 elements and 5! arrangements for those elements in a cycle, noting that 5 orderings yield the same permutation.
  • One participant maintains that the original solution of 504 is correct and does not see any errors in the reasoning presented.
  • A later post introduces a general formula for the order of the centralizer and discusses the conjugacy class order, leading to the same conclusion of 504.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial calculation of 504, with some supporting it and others challenging its validity based on the counting of cycles. The discussion remains unresolved regarding the final answer.

Contextual Notes

There are unresolved assumptions regarding the treatment of cycle equivalences and the implications of the centralizer's order in the context of permutations.

astronut24
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what is the number of elements of order 5 in the permutaion group S7??
so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7 to themselves...so we end up getting 2 cycles of length 1 each. the remaining five elements can be arranged in 4! ways...
so, the answer is 7C2 * 4! = 21*24 = 504.
:smile: but unfortunately, this answer is WRONG!
CAN ANYBODY TELL ME WHY?! PLEEEEASE HELP!
 
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Since 5 is prime, you MUST have cycle of length 5.
Then there could be a 2-cycle in addition.
 
hello

well...yes, it's possible to have a cycle of length 2 in addition to the 5 cycle...but then the l.c.m becomes 10. so that rules out such a consideration!
 
Yes, so the only permutations in S7 of order 5 are the 5-cycles.
 
Your answer is wrong because you've counted, for example:

12345, 23451, 34512, 45123, and 51234 as different elements.
 
ya...so what further? that's a valid point you've raised...
so do you divide by 4?
 
You have [itex]{7 \choose 5}[/itex] ways to pick 5 elements from a set of 7.
There are 5! ways you can order 5 elements in a cycle.
For a given cycle of length 5, 5 orderings are give the same permutation.
 
hey galileo

please read stuff carefully...
we left this a long while ago, right astronut?! :smile:
 
i think the solution 504 is correct...
i don't see any fallacy in it.
 
  • #10
The general solution is:

suppose t is a permutation of type

[tex]1^{m_1}2^{m_2}...r^{m_r}[/tex]

then the order of the centralizer is

[tex]\prod_i i^{m_i}m_i![/tex]

in this case it is 1^2.5

so the centralizer's order is

2!.5

hence the conjugacy class has order

7!/10

which is indeed 504
 

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