Quick method to find conjugates in cycle notation?

Question

Using elements of $S_n$, is there a more efficient method to calculate $x^g = g^{-1}xg$ than just writing out the cycle notation and calculating them??

Answer 1

Conjugation has the effect of applying the cycle $g$ to the representation for $x$, so you don't have to calculate. Here's a random example. Let $n = 6$, and let $x = (1 \ 4 \ 3)(2 \ 5)$. Now suppose we conjugate by $g = (1 \ 2 \ 3)$. This has the effect of cycling all the $1$'s to $2$'s, the $2$'s to $3$'s and the $3$'s to $1$'s. So the result is simply $x^g = (2 \ 4 \ 1)(3 \ 5)$.

You should try to prove this result! It's morally similar to changing basis in linear algebra, if you're familiar with that. One way you might prove it is to actually think of the elements of the symmetric group as particular matrices of 1s and 0s, and studying the change of basis.