Conjugates in symmetric groups

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Homework Help Overview

The discussion revolves around finding the number of conjugates of specific permutations in symmetric groups, specifically (1,2,3,4) in S7 and (1,2,3) in S5. Participants are exploring the relationship between cycle structure and conjugacy in symmetric groups.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the cycle structure of permutations and how it relates to their conjugates. One participant attempts to find a formula for calculating the number of conjugates and questions the use of binomial coefficients. Another participant expresses uncertainty about their calculations regarding the number of conjugates.

Discussion Status

The discussion is ongoing, with participants offering insights into potential formulas and questioning their own calculations. There is recognition of the need to adjust calculations based on the cycle structure and the parameters used.

Contextual Notes

Participants are navigating the complexities of symmetric groups and the definitions of conjugacy, with some uncertainty about the appropriate parameters for their calculations.

kimberu
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Homework Statement


The question is, "How many conjugates does (1,2,3,4) have in S7?

Another similar one -- how many does (1,2,3) have in S5?


The Attempt at a Solution


I know that the conjugates are all the elements with the same cycle structure, so for (123) I found there are 20 conjugates by hand listing them (132)...(354) etc. But I was wondering if there's some equation to find this amount- I haven't taken probability but I think there's got to be a statistical way to figure it out! Thanks a lot.
 
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the binomial coefficient might be what you're looking for.
"nCr"
"choose function"
 
using the formula on (1,2,3) in S5, I got that it has 10 conjugates, which is wrong -- which N and R should I use for this example if not 5 and 3 (or am I calculating wrong)?
 
No, you're doing it right (i.e. "this isn't what you're looking for). I forgot to divide by 2! (!)... :/
 

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