Prove that $\tau (x_1 x_2 ... x_k) \tau^{-1} = (\tau(x_1) \tau(x_2) ... \tau (x_k))$

Question

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.

Answer 1

HINT:

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

Answer 2

Work an example. Suppose $\tau$ is $(12)(34)$ in cycle notation, and you want to conjugate $\sigma = (123)$. The claim is that $\tau \sigma \tau^{-1} = (214)$, applying $\tau$ "inside" $\sigma$. If you work through this example you should both understand the notation and see the proof.