# 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:

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#### 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 $$\tau\sigma\tau^{-1}=\rho$$.