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Let $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{pmatrix}$$ be a permutation so that $$\pi^p = \underbrace{\pi \pi \ldots \pi}_{p \text{ times}} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{pmatrix}$$ How do you find $p$ where $p$ is a natural number different from $0$?

I tried solving this problem by multiplying the permutation by itself a bunch of times until i got the right answer but that just doesn't seem like the smartest way of solving it , there must be another way.

yberman
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Serbacul
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  • The order of a permutation is the least common multiple of the lengths of the cycles appearing in the disjoint cyclic representation of the permutation. In the case that it is a cycle by itself... the order is simply the length of the cycle. – JMoravitz Aug 02 '21 at 12:53
  • @JMoravitz can you explain it in more depth please ? or provide a link to where it is explained best ? – Serbacul Aug 02 '21 at 13:06
  • What aspect of it? That $\sigma^p = e$ when $\sigma$ is a cycle and $p$ is the length of the cycle? Think about it for a moment... $\sigma$ maps the first element in the cycle to the second element, the second element to the third, and so on... If you apply the permutation twice you'll have the first element map to the third as a result and in the end you should be able to easily show then that $\sigma^k$ maps the first element of the cycle to the $(k+1)$'st element of the cycle... wrapping around back to the start as necessary. – JMoravitz Aug 02 '21 at 13:10
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    Think of a dial or a clock or something physical like that... A cyclic permutation in this context would be something like rotating the dial. $\sigma^p=e$ where $p$ is the length of the cycle would correspond to turning the dial all the way around until it is back where it started... and the amount you had to twist it by was the length of the dial itself. – JMoravitz Aug 02 '21 at 13:12
  • As for why the order of a permutation is the least common multiple of the lengths of the cycles of the permutation when written in disjoint cyclic form... see here or here or here etc... – JMoravitz Aug 02 '21 at 13:16
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    oh that was really helpful , thanks for the really in depth explanation and the analogy . And again sorry for bad formatting I am still trying to get the hang of this type of formatting language and cheers once more for answering so fast – Serbacul Aug 02 '21 at 13:16

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