Show that $\tau (18632)(47) \tau^{-1} = (12345)(67)$.

by 72D   Last Updated November 18, 2018 12:20 PM

Example 4.7. of Aluffi's Algebra says

In $S_8$, $(18632)(47)$ and $(12345)(67)$ must be conjugate since they have the same type. So there there exist $\tau$ such that $\tau (18632)(47) \tau^{-1} = (12345)(67)$ for $\tau = (285)(364)$.

If $\tau (18632)(47) \tau^{-1} = (12345)(67)$ means $\tau (18632)(47) = (12345)(67) \tau$ so for RHS applying $\tau$ to $(\text{id})$ then $(12345)(67)$ results in $(128)(3765)$; and for LHS applying $(18632)(47)$ to $(\text{id})$ then $\tau$ results in $(152)(3847)$ which indeed are not equal! Which part am I doing wrong?

Edit. $\tau$ is suggested by the book:

enter image description here



Answers 1


Maybe it is a printing mistake for $\tau$ , I think $\tau$ in the book is actually suppose to be $\tau^{-1}$ and that would work. because in general: If $\sigma=(i_1^1...i_{t(1)}^1)(i_1^2...i_{t(2)}^2)...(i_1^r...i_{t(r)}^r)$ and $\rho=(j_1^1...j_{t(1)}^1)(j_1^2...j_{t(2)}^2)...(j_1^r...j_{t(r)}^r)$, Then define $\tau$ by $\tau(i_b^a)=j_b^a$, then $\tau\sigma\tau^{-1}=\rho$.

mathnoob
mathnoob
November 18, 2018 12:17 PM

Related Questions


Multiplication in permutation Group- cyclic

Updated August 26, 2018 22:20 PM

Abstract Algebra Proof with cycles and transpositions

Updated October 11, 2018 02:20 AM

Artin's Algebra Chapter 1 Exercise 5.5

Updated August 04, 2018 04:20 AM

When do two permutations commute?

Updated August 05, 2018 17:20 PM