# Let f be an r-cycle in $S_n$. Given any $h \in S_n$, show that $hfh^{-1} = (h(x_1), h(x_2), \dots, h(x_r))$

The problem: let $(x_1, x_2, …, x_r)$ be an r-cycle in $S_n$. Show that for every $h \in S_n$, $h \circ (x_1, x_2, \ldots, x_r) \circ h^{-1} = (h(x_1), h(x_2), \ldots, h(x_r))$

#### Solutions Collecting From Web of "Let f be an r-cycle in $S_n$. Given any $h \in S_n$, show that $hfh^{-1} = (h(x_1), h(x_2), \dots, h(x_r))$"

For $i \notin \{h(x_1), \dots, h(x_r)\}$ you have
$$h \circ (x_1 \ \dots \ x_r) \circ h^{-1}(i)= h \circ h^{-1}(i)=i$$

And for $i =h(x_j) \in \{h(x_1), \dots, h(x_r)\}$ you have
$$h \circ (x_1 \ \dots \ x_r) \circ h^{-1}(i)= h \circ (x_1 \ \dots \ x_r)(x_j)=h (x_{j+1})$$