If $\beta = \left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)\left(2\; 4 \;10\right) \in S_{10}$, what is the smallest integer such that $\beta^n = \beta^{-5}$?

Question

I manually found that $\beta^{-5} = \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)\left( 2 \;10 \;4\right)$. The nice thing is that since beta is composed of two disjoint cycles, so I know that $\beta^n = \left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^n\left(2\; 4 \;10\right)^n$. Now I just tried to find an $n$ such that $\left(2\; 4 \;10\right)^n = \left( 2 \;10 \;4\right)$ after finding one I would check to see if $\left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^n= \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)$. Clearly $n=2$ works and luckily $\left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^2= \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)$. So I can conclude that $\beta^2 = \beta^{-5}$.

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My question is how do I check is $n=2$ is indeed the smallest integer. All I showed so far is that $\beta^2 = \beta^{-5}$ and nothing about if 2 is the lowest value. One possibility is to consider the integers from -2 to 1 and try to show that those fail. But is there a better way to do a question like this.

Answer 1

This is the smallest $n$ such that the order of $\beta$ divides $n+5$.

Since you have disjoint cycles of length $7$ and $3$, the order is $21$. Hence, $n = 16$.