I've been given the following question in the context of group actions through conjugation but I'm having difficulty understanding what is being asked
Let $\tau$ be any permutation in $S_m$.
Let $\sigma$ be a cycle $\sigma = (a_1a_2...a_n)$ in $S_m$. Show that $\tau\sigma\tau^{-1}$ takes $\tau(a_1) \rightarrow \tau (a_2)$, $\tau(a_2) \rightarrow \tau (a_3)$, $...$ ,$\tau(a_n) \rightarrow \tau (a_1)$. Hence $\tau\sigma\tau^{-1}=(\tau(a_1)\tau(a_2)...\tau(a_n))$.
I do not quite understand what $\tau(a_1)$ means. To me it seems that $\tau(a_1) = \tau$ since $(a_1)$ is a permutation of 1 item. I'm guessing $ \tau(a_1)$ can be thought of as a function? But I am not quite sure how to intepret this.
Any help would be appreciated.
Regarding the proof (vs.the notation)
Note $A=\{\tau(a_1), \dots ,\tau(a_n)\}$ and $\mathbb N_m=\{1, \dots, m\}$.
For $x \in \mathbb N_m \setminus A$, you have $\tau^{-1}(x) \notin \{a_1, \dots ,a_n\}$. Hence $\sigma \tau^{-1}(x)=\tau^{-1}(x)$ and $x=\tau \sigma \tau^{-1}(x)=(\tau(a_1)\tau(a_2)...\tau(a_n))(x)$ as $x \notin A$.
While for $x=\tau(a_i)$ with $1 \le i \le n$: $$\tau \sigma \tau^{-1}(x)=\tau\sigma(a_i)=\tau(a_{i+1})=(\tau(a_1) \ \dots \ \tau(a_n))(\tau(a_i))$$