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1. Let T and (abc) be permutations in a symmetric group S. Prove thatτ (abc)τ1 = (τ(α) + τ(b) + τ(c)My wrong answerLet T E Sn be any permutation(a, b, c) E S be any 3-cycleProve: T(a b c)r-1 = (t(a) + T(b) r(c))(a, b, c) E Sn meansb, bc, c aSo to showа —aT(a b c)r is the 3 - cycle:τ (α) τ(b), τ ( b) > τ(c), τ(c) > τ(α)CBy changing the labels a, b and c we may assume that the 3-cycle is (123). Now anypermutation t is a product of transpositions, so it is enough to prove the statementwhen T is a transpositon --that is ,T changes only two indices, leaving the others fixedT.Case 1: T does not involve any of the indices 1,2 or 3. In this case T commutes with (123)and there is nothingr =(45)). Note that (45) (45) identityto prove. (we may assume TpermutationT(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)-1Case 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 onlyone of the symbols 1,2 or 3. Assume t -(14). That is, r takes 1 to4 and 4 to 1T(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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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

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
help_outline

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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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Step 1

Given two permutations in a symmetric group are

r and (a b c) e S
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r and (a b c) e S

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Step 2

Consider, the permutation of 3-cycle as

a (a b c) and let r(a) such that 1 sis3
r(a) a and a(a)= a1 md3
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a (a b c) and let r(a) such that 1 sis3 r(a) a and a(a)= a1 md3

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Step 3

Now take j such that j belo...

ja for any i
then a(jjsincej is not in the 3-cycle defining a.
ur"(r(j))= r(5)
So, τατ
Since, taT
fixes any number which is not of the form r(a) for some i.
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ja for any i then a(jjsincej is not in the 3-cycle defining a. ur"(r(j))= r(5) So, τατ Since, taT fixes any number which is not of the form r(a) for some i.

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