- #1
PiRGood
- 14
- 0
Hi all, long time reader first time poster! Just need a hand on this problem I've been stuck on for a few days
Let r=(a_1,a_2...a_k) be in S_n. Suppose that ß is in S_n. Show that:
ßrß^-1=(ß(a_1), ß(a_2)...ß(a_k)).
Basically here i need to prove that the cycle structure of transposition is preserved in composition. For example if r=(1,2)(3,4) and ß=(1,2,3,4) then ßrß^1 = (ß(1),ß(2),ß(3),ß(4))=(2,3)(4,1). That is, r is a double transposition and so is ßrß^1.
I've been able to solve plenty when using examples like the one above but i can't seem to do it without loss of generality. Any help would be greatly appreciated :)
Homework Statement
Let r=(a_1,a_2...a_k) be in S_n. Suppose that ß is in S_n. Show that:
ßrß^-1=(ß(a_1), ß(a_2)...ß(a_k)).
Homework Equations
The Attempt at a Solution
Basically here i need to prove that the cycle structure of transposition is preserved in composition. For example if r=(1,2)(3,4) and ß=(1,2,3,4) then ßrß^1 = (ß(1),ß(2),ß(3),ß(4))=(2,3)(4,1). That is, r is a double transposition and so is ßrß^1.
I've been able to solve plenty when using examples like the one above but i can't seem to do it without loss of generality. Any help would be greatly appreciated :)