1. Let T and (abc) be permutations in a symmetric group S. Prove that τ (abc)τ1 = (τ(α) + τ(b) + τ(c) My wrong answer Let T E Sn be any permutation (a, b, c) E S be any 3-cycle Prove: T(a b c)r-1 = (t(a) + T(b) r(c)) (a, b, c) E Sn means b, bc, c a So to show а — a T(a b c)r is the 3 - cycle: τ (α) τ(b), τ ( b) > τ(c), τ(c) > τ(α) C By changing the labels a, b and c we may assume that the 3-cycle is (123). Now any permutation t is a product of transpositions, so it is enough to prove the statement when T is a transpositon --that is ,T changes only two indices, leaving the others fixed T. Case 1: T does not involve any of the indices 1,2 or 3. In this case T commutes with (123) and there is nothing r =(45)). Note that (45) (45) identity to prove. (we may assume T permutation T(1 2 3)r = (45)(1 2 3)(45) = (45)(45)(1 2 3) = (1 2 3) (T(1)r(2)t(3)) As t fixes (1,2, 3) -1 Case 2: assumeT is a transposition involving 2 of the symbols 1, 2,3. say T (12) T(1 2 3)r = (12)(1 2 3) (12) (132) (213) (T(1)r (2)7(3)) -1 = Case 3: T involves only one of the symbols 1,2 or 3. Assume t -(14). That is, r takes 1 to 4 and 4 to 1 T(1 2 3)r (14) 1 2 3) (14) = (234) = (423) - (τ(1)τ (2)τ (3)) = as t fixes 2,3 and t(1) fixes 4
1. Let T and (abc) be permutations in a symmetric group S. Prove that τ (abc)τ1 = (τ(α) + τ(b) + τ(c) My wrong answer Let T E Sn be any permutation (a, b, c) E S be any 3-cycle Prove: T(a b c)r-1 = (t(a) + T(b) r(c)) (a, b, c) E Sn means b, bc, c a So to show а — a T(a b c)r is the 3 - cycle: τ (α) τ(b), τ ( b) > τ(c), τ(c) > τ(α) C By changing the labels a, b and c we may assume that the 3-cycle is (123). Now any permutation t is a product of transpositions, so it is enough to prove the statement when T is a transpositon --that is ,T changes only two indices, leaving the others fixed T. Case 1: T does not involve any of the indices 1,2 or 3. In this case T commutes with (123) and there is nothing r =(45)). Note that (45) (45) identity to prove. (we may assume T permutation T(1 2 3)r = (45)(1 2 3)(45) = (45)(45)(1 2 3) = (1 2 3) (T(1)r(2)t(3)) As t fixes (1,2, 3) -1 Case 2: assumeT is a transposition involving 2 of the symbols 1, 2,3. say T (12) T(1 2 3)r = (12)(1 2 3) (12) (132) (213) (T(1)r (2)7(3)) -1 = Case 3: T involves only one of the symbols 1,2 or 3. Assume t -(14). That is, r takes 1 to 4 and 4 to 1 T(1 2 3)r (14) 1 2 3) (14) = (234) = (423) - (τ(1)τ (2)τ (3)) = as t fixes 2,3 and t(1) fixes 4
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I need help understanding attached question for abstract algebra. My answer teacher said dosnt accountt for all numbers that may be in a set, i only used up to 5. Please help and show me correct way
Transcribed Image Text:1. Let T and (abc) be permutations in a symmetric group S. Prove that
τ (abc)τ1 = (τ(α) + τ(b) + τ(c)
My wrong answer
Let T E Sn be any permutation
(a, b, c) E S be any 3-cycle
Prove: T(a b c)r-1 = (t(a) + T(b) r(c))
(a, b, c) E Sn means
b, bc, c a
So to show
а —
a
T(a b c)r is the 3 - cycle:
τ (α) τ(b), τ ( b) > τ(c), τ(c) > τ(α)
C
By changing the labels a, b and c we may assume that the 3-cycle is (123). Now any
permutation t is a product of transpositions, so it is enough to prove the statement
when T is a transpositon --that is ,T changes only two indices, leaving the others fixed
T.
Case 1: T does not involve any of the indices 1,2 or 3. In this case T commutes with (123)
and there is nothing
r =(45)). Note that (45) (45) identity
to prove. (we may assume T
permutation
T(1 2 3)r
= (45)(1 2 3)(45) = (45)(45)(1 2 3)
= (1 2 3) (T(1)r(2)t(3))
As t fixes (1,2, 3)
-1
Case 2: assumeT is a transposition involving 2 of the symbols 1, 2,3. say T (12)
T(1 2 3)r
= (12)(1 2 3) (12) (132) (213)
(T(1)r (2)7(3))
-1
=
Case 3: T involves only
one of the symbols 1,2 or 3. Assume t -(14). That is, r takes 1 to
4 and 4 to 1
T(1 2 3)r
(14) 1 2 3) (14) = (234) = (423)
- (τ(1)τ (2)τ (3))
= as t fixes 2,3 and t(1) fixes 4
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