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$S_7$, the symmetric group on 7 letters.

Find the order of $\pi$ where

$$ \pi = \begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 2 & 3 & 7 & 5 & 1 & 4 & 6 \end{bmatrix}$$

its 7 yes?? I am trying to do the permutations but any element of the group is suppose to divide the order of $S_7$?

I am doing the permutations on scratch paper. I am on $\pi^4=$ and its not looking like $e$

Work in progress

computing $\pi * \pi $

$$ \begin{aligned} \pi * \pi =\begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 2 & 3 & 7 & 5 & 1 & 4 & 6 \end{bmatrix} *\begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 2 & 3 & 7 & 5 & 1 & 4 & 6 \end{bmatrix} \end{aligned}$$

$$ \begin{aligned} 1 \to 2 \to 3 \\ 2 \to 3 \to 7 \\3 \to 7 \to 6 \\4 \to 5 \to 1 \\5 \to 1 \to 2 \\6 \to 4 \to 5 \\7 \to 6 \to 9 \end{aligned}$$

making $$ \pi^2= \begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 3 & 7 & 6 & 1 & 2 & 5& 4 \end{bmatrix} $$

computing $\pi^3 $

$$ \pi^3 =\pi^2 * \pi = \begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 7 & 6 & 4 & 2 & 3 & 1 & 5 \end{bmatrix} \begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 2 & 3 & 7 & 5 & 1 & 4 & 6 \end{bmatrix} $$

$$ \begin{aligned} 1 \to 2 \to 7 \\ 2 \to 3 \to 6 \\3 \to 7 \to 4 \\4 \to 5 \to 2 \\5 \to 1 \to 3 \\6 \to 4 \to 1 \\7 \to 6 \to 5 \end{aligned}$$

making $$ \pi^3= \begin{bmatrix} 1& 2 & 3 & 4& 5 & 6 & 7 \\ 7 & 6 & 4 & 2 & 3 & 1 & 5 \end{bmatrix} $$

Tiger Blood
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  • Yes, it's order 7. Check out https://math.stackexchange.com/questions/31763/multiplication-in-permutation-groups-written-in-cyclic-notation , may be helpful to you – Theo C. Sep 27 '17 at 01:26
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    The order of $S_7$ is $7!$, and $7$ certainly divides $7!$. – Joppy Sep 27 '17 at 01:26
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    "I am on $\pi^4$ and it's not looking like $e$" . Of course not - keep going! – NickD Sep 27 '17 at 01:35

2 Answers2

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A quicker method would be to use the one line cycle notation for $\pi$, $$\pi = (1\;2\; 3\; 7\; 6\; 4\; 5).$$ This is just one 7-cycle, so the order is 7.

videlity
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"Order of permutation of a finite set in disjoint cycles is the least common multiple of length of cycles." As you can see The permutation here has $$\pi=(1 2 3 7 4 6 5 )$$ which is the only disjoint cycle ,and of length 7 . Hence the order of $\pi$ is 7 .

  • You just answered using the same argument as the already accepted answer. – mucciolo Nov 05 '17 at 21:20
  • @mucciolo Pardon me sir ,but I thought "with a theorem" answer would be better for upcoming readers of the question . –  Nov 05 '17 at 21:23