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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.

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    $\begingroup$ $\tau$ is some permutation. $\tau (a_1)$ would mean the element to which $a_1$ is mapped to by $\tau$. $\endgroup$
    – vnd
    Oct 18, 2015 at 12:49
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    $\begingroup$ In this case, these are "function application parentheses". $\tau(a_1)$ is just the value of "$\tau$ evaluated at $a_1$". $\endgroup$ Oct 18, 2015 at 12:49

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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 $1 \le i \le n$: $$\tau \sigma \tau^{-1}(x)=\tau\sigma(a_i)=\tau(a_{i+1})=(\tau(a_1) \ \dots \ \tau(a_n))(\tau(a_i))$$

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