Prove that there is no a such that $a^{-1}(1,2,3)a = (1,3)(5,7,8)$

Question

Herstein 2.10.8 (b)

Prove that there is no a such that $a^{-1}(1,2,3)a = (1,3)(5,7,8)$

My initial thought was that order of RHS and LHS permuation should be same.

For RHS, the cycles are disjoint, so order = lcm(2,3) = 6

And for LHS, I got to know that it has order 3. And thus there is no such 'a' possible.

What I don't understand is how did we come to conclusion that $a^{-1}(1,2,3)a$ has order 3 ? Clearly (1,2,3) has order 3, but after taking product with some other permutation, will the order remain same, even if the cycles are not disjoint ?

Answer 1

In any group $(x^{-1}gx)^{n}=x^{-1}g^n x$ and so $g^n=e$ if and only if $x^{-1} g^n x=e$ if and only if $(x^{-1}gx)^{n}=e$.

Answer 2

$a^{-1}pa$ must yield an element in the same conjugacy class as $p$, and conjugate elements have the same order.