A question concerning 4-cycles in $S_4$

Question

Is it true that for all $g\in S_4$ and $f \in S_4$ a 4-cycle, then $g^{-1}fg=h$ implies $h$ is also a 4 cycle. I did a few examples, and it seems to be true, but I don't know how to prove it. Also, I guess it works in $S_n$ as well.

Answer 1

Yes, if $f \in S_4$ is a 4-cycle and $g \in S_4$, then $g^{-1}fg$ is also a 4-cycle. More generally, if $f, g \in S_n$, then $g^{-1}fg$ has the same cycle structure as $f$, and the permutation $g^{-1}fg$ can be obtained by replacing the element $i$ in $f$ by $g(i)$. For example, if $f=(1,2,4)(3,6)(5)$ and $g \in S_6$, then $g^{-1}fg$ is the permutation $(g(1), g(2), g(4))(g(3),g(6))(g(5))$, which has the same cycle structure as $f$. The proof is simple: Suppose $f$ takes $i$ to $j$. For conciseness let $f(i)$ be denoted by $i^f$. Then $g^{-1}fg$ takes $g(i)$ to $g(i)^{g^{-1}fg} =i^{fg} = j^g = g(j)$.