Prove that $\tau (x_1 x_2 ... x_k) \tau^{-1} = (\tau(x_1) \tau(x_2) ... \tau (x_k))$ I want to prove that $\forall \tau \in S_n$ and for pairwise different $x_1,...,x_k \in [n]$ it holds true that:
$\tau (x_1 x_2 ... x_k) \tau^{-1} = (\tau(x_1) \tau(x_2) ... \tau (x_k))$
I don't quite understand the notation. I assume that $S_n$ is the symmetric group and $\tau$ is any permutation in $S_n$ and $(x_1 x_2 ... x_k)$ is a cycle of length $k$. However I don't understand what the second part of the equation: $(\tau(x_1) \tau(x_2) ... \tau (x_k))$ means.

HINT:
What is $\tau(x_1 x_2 \ldots x_k)\tau^{-1}$ applied to $\tau(x_1)$ ?
