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I am making a mistake in solving the following problem. I would really appreciate it if someone looked over it and helped me with the solution

$a=(1,2,3)(2,3,4)(5,6,7)(7,8,9,10)$

I need to show that the order of this permutation is 10. What I did was, I tried to write the permutation as a product of disjoint cycles and find the least common multiple of the lengths of them. So, the computation yielded:

$a=(7,8,9,10,5,6)(4,3)(2,1)$

The order of this permutation is not 10.

I am making some mistake writing the permutation as disjoint cycles, but I cannot find where

Help would be appreciated!

Thanks in advance!

P.S. Does $10$ go to $5$?

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  • $\begingroup$ 10 does go to 5 $\endgroup$ – Shuri2060 Mar 19 '16 at 20:40
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$a=(1,2,3)(2,3,4)(5,6,7)(7,8,9,10)$

This means that $(1,2,3,4,5,6,7,8,9,10)$ is permuted first by $(7,8,9,10)$, then $(5,6,7)$, then $(4,5,6)$ and then $(1,2,3)$.

To deal with this, look at each number separately and start with their appearance on the right to see where they map to.

If we start with 1, for example, its first appearance is in the fourth permutation. So it maps to 2.

On the other hand, if we start with 3, its first appearance is in the third permutation, $(2,3,4)$. So it maps to 4. 4 does not appear in the next permutations so we move on.

EDIT: I've just used this: which agrees with your answer. (Note you can use letters in this calculator and * to compose permutations)

So it was probably a mistake in the question.

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  • $\begingroup$ Thats what I am doing too, man, I know the procedure $\endgroup$ – Frank Booth Mar 19 '16 at 20:41
  • $\begingroup$ Briefly checking your answer, I think I agree. Where do you get that its order is 10? It could be a mistake in the question. $\endgroup$ – Shuri2060 Mar 19 '16 at 20:45
  • $\begingroup$ It says, show that $a$ has order 10 in $S_n, n\ge 10$ $\endgroup$ – Frank Booth Mar 19 '16 at 20:48
  • $\begingroup$ Is it a textbook? $\endgroup$ – Shuri2060 Mar 19 '16 at 20:48
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    $\begingroup$ I've just used this: bananattack.com/permutations which agrees with your answer. (Note you can use letters in this calculator and * to compose permutations) $\endgroup$ – Shuri2060 Mar 19 '16 at 20:50

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