We have: $\alpha = (a_1a_2 \cdots a_r), \beta=(b_1b_2\cdots b_r)\in S_n$ ($\alpha,\beta$ are strange cycles) How can we find $f\in S_n$ s.t.: $$\beta=f\alpha f^{-1}\;\;\;?$$

Thank you!

(The answer should be as multiply of strange permutation cycles...)

strange = disjoint

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    $\begingroup$ When you say strange, do you mean disjoint? (This looks like a translation problem.) $\endgroup$ – Brian M. Scott Nov 24 '13 at 19:07
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    $\begingroup$ Also a further clarification that might be good to have: Do you in that case mean that $\alpha$ and $\beta$ are disjoint (this is the only interpretation I can think of, but it will not actually change the answer, so it seems strange if an exercise would contain that condition). $\endgroup$ – Tobias Kildetoft Nov 24 '13 at 19:17
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    $\begingroup$ Just find an $f$ that maps each $a_i$ to $b_i$. As Tobias Kildetoft says, there is no need to assume that the cycles are disjoint. $\endgroup$ – Derek Holt Nov 24 '13 at 19:56

The key to answering this question is knowing that

$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)).$$

To see this, note that any $b \in \{1, \dots, n\}$ is equal to $f(a)$ for some $a$. If $a \notin \{a_1, \dots, a_r\}$, then

$$(f\alpha f^{-1})(f(a)) = f(\alpha(a)) = f(a).$$ If $a = a_i$, then $$(f\alpha f^{-1})(f(a_i)) = f(\alpha(a_i)) = f(a_{i+1})$$ where $a_{r+1}$ should be interpreted as $a_1$. Therefore,

$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)).$$

With this fact at your disposal, you want to find $f$ such that

$$f\alpha f^{-1} = (f(a_1)\, \dots\, f(a_r)) = (b_1\, \dots\, b_r).$$

As Derek Holt suggested in the comments, you can take $f(a_i) = b_i$ and extend to a bijection. Note, this is not the only such $f$.

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