Permutations (last question of sheet, yay )

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SUMMARY

The discussion focuses on determining the sign of a specific permutation in the symmetric group Sn and demonstrating a property of cycles under conjugation. The sign of the permutation is established as (-1)n/2 for even n and (-1)(n-3)/2 for odd n. The user seeks clarification on handling negative numbers and rounding for odd values of n. Additionally, they inquire about a more concise method for distinguishing between even and odd cases in their calculations.

PREREQUISITES
  • Understanding of symmetric groups, specifically Sn.
  • Knowledge of permutation sign calculations.
  • Familiarity with cycle notation in group theory.
  • Basic algebraic manipulation of exponents and integer properties.
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  • Research the properties of permutation signs in group theory.
  • Study the concept of conjugation in symmetric groups.
  • Learn about cycle decomposition and its implications in Sn.
  • Explore methods for handling odd and even integers in mathematical proofs.
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Students of abstract algebra, mathematicians focusing on group theory, and anyone interested in advanced permutation concepts and their applications.

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Permutations (last question of sheet, yay!)

1. Homework Statement [/b]

\eta:=
(1 2 ... n-1 n)
(n n-1 ... 2 1)
\inS_{n} for any n\inN
n.b That should be 2 lines all in one large bracket btw
a.) Determine its sign.

b.) Let n \geq1. Let <a1,...,as> \inSn be a cycle and let \sigma\inSn be arbitrary. Show that

\sigma\circ <a1,...,as> \circ\sigma^{-1} = <\sigma(a1),...,\sigma(as)> in Sn.

Homework Equations





The Attempt at a Solution



I get the sign of the permutation to be (-1)^n/2

I don;t know how to do the second part, any ideas?
 
Last edited:
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Actually, i thought i had done the first part, but i haven't because I am stuck on how to show that for negative numbers, i want the n/2 to be taken as the rounded down value. For example if n=7 i want n/2 to be taken as 3. Is there is a simple way to do this for odd numbers but also keep the same form for positive values of n.

Also it should be (-1)^(n-2/2)

So it should be (-1)^(n-2/2) for even numbers of n and (-1)^(n-3/2) for odd values of n, is there a neater way to do this?
 
Last edited:
I have just said for even numbers of n, that is, n/2 \inZ and for odd numbers, that is n/2\notinZ to distinguish between the two cases.

Im still thinking through part b.) so any help is welcomed.
 

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