If $y$ is a cycle of length $r$, show that $\sigma y \sigma^{-1}$ is also a cycle of length $r$. I am trying to show that if $y$ is a cycle of length $r$, and $\sigma \in S_n$ then $\sigma y \sigma^{-1}$ is also a cycle of length $r$. More specifically, that if $y = (k_1\ \dots k_r)$ then $\sigma y \sigma^{-1} = (\sigma(k_1) \dots \sigma(k_r))$
I am not too sure how to show this. I know that $\sigma$ can be written as the product of disjoint cycles or length at least 2 by the cycle decomposition theorem, but am not sure how that helps. 
 A: HINT: For convenience let $x=\sigma y\sigma^{-1}$. Suppose that $y(i)=j$, where we allow $j=i$; then $x\big(\sigma(i)\big)=\sigma\big(y(i)\big)=\sigma(j)$. Thus, if $i$ is a fixed point of $y$, then $\sigma(i)$ is a fixed point of $x$. And if $i$ and $j$ are consecutive members of the cycle $y$, then ... ?
A: *

*First justify that $\sigma y\sigma^{-1}$ will send $\sigma(k_i)$ on $\sigma(k_{i+1})$ if $i<r$ and will send $\sigma(k_r)=\sigma(k_1)$.

*Second, take $t\notin\{\sigma(k_1),...,\sigma(k_r)\}$ then justify that $\sigma y\sigma^{-1}$ will send $t$ on $t$ (Hint : if $t\notin\{\sigma(k_1),...,\sigma(k_r)\}$ then $\sigma^{-1}(t)\notin\{k_1,...,k_r\}$ and hence, $\sigma^{-1}(t)$ is fixed by $y$).

*$1$ and $2$ imply that $\sigma y\sigma^{-1}=(\sigma(k_1),...,\sigma(k_r))$.
A: Incomplete answer: Conjugation preserves order (whether or not this is proved because conjugation is an automorphism, and automorphisms preserve order), and order of a cycle equals its length. Therefore, the conjugate of a cycle has the same order as the cycle. If the conjugate of a cycle is a cycle, then since it has the same order, it has the same length.
