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Let $\beta$, and $\alpha = \{x_1,x_2, \cdots,x_{r-1}, x_r\}$, be permutations of a finite set X. Show that $$\beta \circ \alpha \circ \beta^{-1}=(\beta(x_1) \space \beta(x_2) \space \cdots \space \beta(x_{r-1}) \space \beta(x_r)).$$ We haven't formally learned modular arithmetic yet so the teacher gave us the hint to write a function defined as follows, $T(x_i)=x_{i+1}$ if $x_i \neq r$, If $x=i$ then $T(x_i)=x_1$.

Then we consider the $2$ cases being if $x_i \in \beta$ and when $x_i \notin \beta$.
I'm still not completely sure how to procede down either case, any help is appreciated!

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Hint. We can assume that $\alpha$ is a cycle; we have $\alpha(x_1)=x_2$ and so on. We have to show that $\beta\alpha\beta^{-1}$ maps $\beta(x_1)$ to $\beta(x_2)$, and so on: we have $$(\beta\alpha\beta^{-1})(\beta(x_1)) =\beta(\alpha(\beta^{-1}(\beta(x_1))))=\beta(\alpha(x_1))=\beta(x_2)$$ as required. See if you can do the rest yourself.

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    Then for the proof I would say $\alpha(x_i)=x_{i+1}$ and do exactly as you did to show $\beta(x_i)$ maps to $\beta(x_{i+1})$. Then as a separate case consider when $\alpha(x_r)=x_1$ – 2017-02-20
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    And also consider what happens if $x$ is unchanged by $\alpha$: show it is also unchanged by $\beta\alpha\beta^{-1}$. – 2017-02-20
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    Ahh yes okay thanks, very helpful! – 2017-02-20