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How can i define the order of a permutation without doing the permutation again and again? Example: say $σ=(1-->2,2-->3,3-->5,4-->1,5-->4)$ in $S_5$.

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Some friendly advice: do not ask a second question when you should ask a *new*@ question (new post). It makes the answers you received seem incomplete. If you have questions about how to find the inverse of a permutation, you need to first search the site to find one of the many duplicate questions on that, or, if you cannot find a duplicate, post it as a new question. – amWhy Jun 24 '14 at 16:24
Im really sorry for the re editing. – Manolis Lyviakis Jun 24 '14 at 16:58
No problem, Manolis. As I said, I was offering some friendly advice. – amWhy Jun 24 '14 at 16:59
Advice is taken.I will do that. – Manolis Lyviakis Jun 24 '14 at 17:01
up vote 3 down vote accepted

Write the permutation in cycle notation (and expressed as the product of disjoint cycles).

For each cycle in the permutation determine its length. (If there is only one cycle of length greater than $1$, then the order of the permutation is the length of that cycle. If all cycles in a permutation are of length one, it is necessarily the identity permutation, and has order $1$.

Calculate the least common multiple of all cycle lengths.

The result is the order of the permutation.

Example: Let $\tau =\begin{pmatrix} 1& 2& 3& 4& 5\\ 2& 4 & 5 & 1& 3\end{pmatrix}$

Written as the product of disjoint cycles gives us $\tau = (1, 2, 4)(3, 5) \in S_5$.

Then the lengths of the cycles, from left to right, are $3, 2$.

$\operatorname{lcm}(3, 2) = 6$.

The order of $\tau$ is equal to $6$.

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The order is the least common multiplier of the length of all circles in the circle notation of the permutation.

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you mean lcm of the lengths of all the "cycles" in the "disjoint cycle" representation of the permutation. – Anurag A Jun 24 '14 at 15:43
Can you o an example pease. Say $S5$ $$σ=(1-->2,2-->3,3,-->5,4-->1,5-->4)$$ – Manolis Lyviakis Jun 24 '14 at 15:46
@manolis that has one cycle of length 5, so the order is 5. – Peter Jun 24 '14 at 15:48
The cycle notation is (12354) in this case. – Peter Jun 24 '14 at 15:48

The element that you have listed isn't permutation because it sends both $4$ and $5$ to $1$. If, however, you had $\sigma = (1235) \in S_5$, then you can find the order by simply multiplying the permutation by itself: $$\begin{align} \sigma^1 &= (1235) \\ \sigma^2 &= (13)(25) \\ \sigma^3 &= (1532) \\ \sigma^4 &= (1). \end{align} $$ So the order of $\sigma$ is $4$.

This, obviously, only works when the group is relatively small.

If you wanted to ask about the order of $\sigma = (12354)$, then just star $$\begin{align} \sigma^1 &= (12354) \\ \sigma^2 &= (13425) \\ \sigma^3 &= \dots. \end{align} $$

In case you are not familiar with the notation, $(1235)$ is the permutation that $$\begin{align} 1 &\longrightarrow 2 \\ 2 &\longrightarrow 3 \\ 3 &\longrightarrow 5 \\ 5 &\longrightarrow 1\\ 4 &\longrightarrow 4. \end{align} $$

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sorry i meant 5-->3 – Manolis Lyviakis Jun 24 '14 at 15:58
@ManolisLyviakis: It still doesn't work. – Thomas Jun 24 '14 at 16:02
******5--->4 sorry – Manolis Lyviakis Jun 24 '14 at 16:22

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